JC2 H2 Maths Bukit Timah | Committee Combinations

Did you know? Choosing a committee is different from arranging people in seats. Suppose five students belong to Group A and four to Group B, and you need three people with at least one from each group. There are C(9,3)=84 unrestricted committees. Remove the ten committees selected entirely from A and the four selected entirely from B. The answer is 84−10−4=70 committees. No extra factor of 3! is needed, because the committee has no ranked positions.

This JC2 H2 Mathematics Tutor Bukit Timah guide explains the exact A-Level counting question: how do you select a committee containing at least one member from each of two distinct groups, without counting it twice or confusing combinations with permutations? The complement method comes first, followed by the direct case-splitting method, worked examples, validity checks and 36 original selection clinics. The examples use hypothetical group sizes, not actual school or tuition enrolments.

The official SEAB H2 Mathematics 9758 syllabus for the 2027 examination includes permutations, combinations, addition and multiplication counting principles, and restrictions. This article owns the “at least one from each group” committee-selection intent. Other guides cover circular seating and repeated-letter arrangements so the same search question is not answered in several indistinguishable articles.

Quick answer: total committees minus the ones missing a group

Suppose there are a distinct members of Group A and b distinct members of Group B, with the groups disjoint, and select an unordered committee of exactly k people. The unrestricted number is C(a+b,k). Exclude committees from A only, C(a,k), and committees from B only, C(b,k). For k≥1 those two excluded events are disjoint, so the number containing at least one from each group is C(a+b,k)−C(a,k)−C(b,k), with C(n,r)=0 when r>n.

Step one: check that the groups are disjoint and that the people are individually distinct.
Step two: verify that k≤a+b; otherwise no committee of that size exists.
Step three: identify whether order or job titles matter. For an ordinary unranked committee, use combinations.
Step four: subtract the all-A and all-B selections from the total.
Step five: independently count the possible splits, j from A and k−j from B.
Step six: check that the answer is nonnegative and cannot exceed C(a+b,k).

Worked example one: five in A, four in B, choose three

With a=5, b=4, k=3, unrestricted committees number C(9,3)=9×8×7/(3×2×1)=84. A-only committees number C(5,3)=10. B-only committees number C(4,3)=4. Therefore the required count is 84−10−4=70. Every remaining committee contains at least one person from each group.

Check by direct cases. A committee can contain exactly one A and two B, giving C(5,1)C(4,2)=5×6=30. Or it can contain exactly two A and one B, giving C(5,2)C(4,1)=10×4=40. Add those disjoint cases: 30+40=70. The two counting methods agree and independently verify the answer.

The direct case-splitting formula

For at least one member of each group, let j be the number selected from A. It must satisfy 1≤j≤a and 1≤k−j≤b. The direct number is Σ C(a,j)C(b,k−j) over valid integers j. The cases are mutually exclusive because a committee cannot simultaneously contain different counts j of Group A members. That is why their counts are added, not multiplied across separate cases.

The two methods count the same object from different directions. The complement method removes forbidden one-group committees from all possible committees; the direct method builds only allowed group compositions. For a small k such as three or four, the direct method is easy to audit. For larger k and two groups, the complement method is often shorter. Use whichever gives a clear explanation without omitting possible cases.

Worked example two: six from A, four from B, choose five

Here a=6, b=4 and k=5. The unrestricted number is C(10,5)=252. All-A committees are C(6,5)=6, while an all-B committee of five is impossible because B has only four people; therefore C(4,5)=0. The required number is 252−6−0=246. Do not treat C(4,5) as a negative number or a factorial with an invalid argument.

A direct check considers possible A counts 1,2,3 and 4. The four cases contribute C(6,1)C(4,4)=6, C(6,2)C(4,3)=60, C(6,3)C(4,2)=120, and C(6,4)C(4,1)=60. They total 246. Every number comes from selecting a subset, not ordering its members around a table.

Worked example three: four from A, two from B, choose four

With a=4, b=2 and k=4, the total number of committees is C(6,4)=15. An all-A committee is possible in exactly C(4,4)=1 way, while an all-B committee of four is impossible. Therefore 15−1=14 committees contain at least one from each group.

Directly, a committee may contain three A and one B: C(4,3)C(2,1)=4×2=8. Or it may contain two A and both B members: C(4,2)C(2,2)=6×1=6. Total 8+6=14. The all-B case is absent because B contains only two people. A correct case split must respect the actual available membership sizes.

Sometimes the restriction is automatic

Suppose a=6, b=6 and k=7. Any seven-person committee must include at least one member from each group, because neither group contains seven people on its own. The total number is C(12,7)=792, and both excluded counts C(6,7) are zero. The formula therefore returns all 792 committees, exactly as the logical reasoning predicts.

This is a useful shortcut before any factorial calculation. If k is larger than both a and b but no larger than their sum, a committee cannot be selected entirely from either group. The at-least-one-each condition then imposes no additional restriction. Recognising a condition that is automatically satisfied can save several unnecessary cases.

At least one from A is not the same as at least one from both

If a question asks only for at least one from A, the forbidden committees are those with no A members, which are all-B selections. The count is C(a+b,k)−C(b,k). There is no need to exclude all-A committees, because an all-A committee already satisfies “at least one from A”. Subtracting both groups in that problem would answer a stronger and different requirement.

For the five-A, four-B, three-person example, “at least one from A” gives 84−C(4,3)=80. “At least one from each” gives 70. “Exactly one from A” gives C(5,1)C(4,2)=30. These are three distinct event descriptions. The words “at least”, “exactly” and “each” carry mathematical information and should be underlined before calculating.

Combinations are not permutations

A committee of three consisting of named people A, B and C is the same committee whether someone writes A,B,C or C,A,B. Counting all six orderings would overcount an unranked selection. Thus C(n,k) is appropriate. But if the question asks for a chairperson, secretary and treasurer with distinct roles, assigning those roles changes the outcome and an arrangement or further factorial factor may be needed.

Always determine what the finished object is. A three-person committee is a set; a list of three named job holders is an ordered assignment. If roles are later assigned to a selected three-person committee, the total can sometimes be the committee count times 3! provided every committee can fill the three distinct roles in all six ways. Do not multiply by 3! automatically when no roles were requested.

The complementary cases are disjoint for nonempty committees

When k≥1, a committee cannot be selected entirely from A and entirely from B at the same time if the groups are disjoint. Therefore the all-A and all-B forbidden categories do not overlap, and subtracting both is valid. If a problem involves three groups or overlapping category labels, the forbidden categories may overlap, requiring more careful inclusion–exclusion.

For example, with three disjoint groups A,B,C, “missing A” and “missing B” can both hold if a committee is chosen entirely from C. Subtracting each missing-group category once would then remove all-C committees too many times. This article’s two-group formula should not be copied blindly into a three-group question without adjusting the counting logic.

Another interpretation trap: groups must be truly disjoint

Suppose a person belongs to both the debate club and the science club. If a committee must include at least one member from each club, a single dual-club member might satisfy both requirements. Then the simple formula with a+b total people double-counts overlapping membership and is not valid. The groups used in this article are explicitly disjoint categories of distinct people.

For overlapping categories, count the actual unique population and classify members into A-only, B-only and both before applying inclusion–exclusion or direct cases. A mathematically correct solution starts from the real eligibility structure in the question. Using a formula derived for disjoint groups without checking that assumption is a model error, not just an arithmetic mistake.

The binomial coefficient has a natural zero convention

When k>a, C(a,k)=0 because there is no way to choose more than a distinct people from a group of size a. This convention avoids inventing negative counts or undefined factorials in the complementary formula. A calculator may reject a command with an impossible selection size, so the learner should interpret it as zero in the combinatorial count rather than forcing the arithmetic.

Similarly, if k>a+b, the total C(a+b,k)=0, because there are not enough people to make the committee. If k=1, a committee cannot include a person from each of two disjoint groups, so the required number is zero. These boundary cases are part of understanding the problem rather than special tricks to memorise.

How to verify the result by a probability argument

If a committee of size k is chosen uniformly from all C(a+b,k) possible unranked committees, the probability that it includes at least one from each group is the favourable count divided by the total: 1−[C(a,k)+C(b,k)]/C(a+b,k), provided 1≤k≤a+b. For the five-A, four-B, choose-three example, this is 70/84=5/6. This probability calculation is a consequence of the counting result, not a different selection method.

The probability can also be checked with the direct favourable count. A student who obtains a ratio greater than one has counted favourable committees more than once or made a combination error. The denominator is the number of equally likely full committees, not the total number of individuals a+b. Be clear about whether the question requests a count or a probability.

One-minute Bukit Timah diagnostic

Ask for five from A, four from B, choosing three. The unrestricted total is 84. The excluded one-group cases are 10 and 4, leaving 70. Then change the requirement to “at least one from A” and obtain 80. Change it to “exactly one from A” and obtain 30. Finally change the committee to named roles: the counting object is no longer the same unranked set. If the learner explains each change, the complement logic is secure.

Thirty-six original committee-selection clinics

Each clinic below uses a disjoint two-group population of distinct people and an unranked committee of size k. The exact count is calculated using both the complement formula and the direct sum over valid group splits; the two methods are checked to agree. Some cases make a one-group committee impossible, so the restriction becomes automatically satisfied or partly automatic. These are original fictional selection exercises, not actual school committee rosters.

1. Group A has 3, Group B has 7, choose 8

Question. A fictional activity has 3 distinct members in Group A and 7 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 8 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(10,8)=45 unrestricted committees. The all-A count is C(3,8)=0; the all-B count is C(7,8)=0. Excluding both forbidden one-group categories gives 45−0−0=45 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(3,1)C(7,7)=3; C(3,2)C(7,6)=21; C(3,3)C(7,5)=21. Adding the contributions gives 45, matching the complement result. For comparison, “at least one from A” alone would allow 45 committees, while “exactly one from A” would allow 3. Use combinations because the internal order of a committee does not matter. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

2. Group A has 4, Group B has 4, choose 2

Question. A fictional activity has 4 distinct members in Group A and 4 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 2 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(8,2)=28 unrestricted committees. The all-A count is C(4,2)=6; the all-B count is C(4,2)=6. Excluding both forbidden one-group categories gives 28−6−6=16 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(4,1)C(4,1)=16. Adding the contributions gives 16, matching the complement result. For comparison, “at least one from A” alone would allow 22 committees, while “exactly one from A” would allow 16. The all-A and all-B excluded categories are disjoint for a nonempty committee drawn from disjoint groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

3. Group A has 5, Group B has 11, choose 6

Question. A fictional activity has 5 distinct members in Group A and 11 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(16,6)=8,008 unrestricted committees. The all-A count is C(5,6)=0; the all-B count is C(11,6)=462. Excluding both forbidden one-group categories gives 8,008−0−462=7,546 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(5,1)C(11,5)=2,310; C(5,2)C(11,4)=3,300; C(5,3)C(11,3)=1,650; C(5,4)C(11,2)=275; C(5,5)C(11,1)=11. Adding the contributions gives 7,546, matching the complement result. For comparison, “at least one from A” alone would allow 7,546 committees, while “exactly one from A” would allow 2,310. When k exceeds a group’s size, selecting k entirely from that group contributes zero. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

4. Group A has 6, Group B has 8, choose 3

Question. A fictional activity has 6 distinct members in Group A and 8 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 3 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(14,3)=364 unrestricted committees. The all-A count is C(6,3)=20; the all-B count is C(8,3)=56. Excluding both forbidden one-group categories gives 364−20−56=288 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(6,1)C(8,2)=168; C(6,2)C(8,1)=120. Adding the contributions gives 288, matching the complement result. For comparison, “at least one from A” alone would allow 308 committees, while “exactly one from A” would allow 168. The direct split method counts j from A and k−j from B, with j restricted to feasible integers. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

5. Group A has 7, Group B has 5, choose 8

Question. A fictional activity has 7 distinct members in Group A and 5 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 8 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(12,8)=495 unrestricted committees. The all-A count is C(7,8)=0; the all-B count is C(5,8)=0. Excluding both forbidden one-group categories gives 495−0−0=495 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(7,3)C(5,5)=35; C(7,4)C(5,4)=175; C(7,5)C(5,3)=210; C(7,6)C(5,2)=70; C(7,7)C(5,1)=5. Adding the contributions gives 495, matching the complement result. For comparison, “at least one from A” alone would allow 495 committees, while “exactly one from A” would allow 0. A numerical answer larger than the unrestricted total signals a counting error. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

6. Group A has 8, Group B has 2, choose 5

Question. A fictional activity has 8 distinct members in Group A and 2 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 5 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(10,5)=252 unrestricted committees. The all-A count is C(8,5)=56; the all-B count is C(2,5)=0. Excluding both forbidden one-group categories gives 252−56−0=196 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(8,3)C(2,2)=56; C(8,4)C(2,1)=140. Adding the contributions gives 196, matching the complement result. For comparison, “at least one from A” alone would allow 252 committees, while “exactly one from A” would allow 0. Changing “at least one from each” into “exactly one from A” changes the permitted split cases. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

7. Group A has 9, Group B has 9, choose 2

Question. A fictional activity has 9 distinct members in Group A and 9 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 2 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(18,2)=153 unrestricted committees. The all-A count is C(9,2)=36; the all-B count is C(9,2)=36. Excluding both forbidden one-group categories gives 153−36−36=81 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(9,1)C(9,1)=81. Adding the contributions gives 81, matching the complement result. For comparison, “at least one from A” alone would allow 117 committees, while “exactly one from A” would allow 81. No extra k! factor belongs in an unranked committee selection. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

8. Group A has 10, Group B has 6, choose 7

Question. A fictional activity has 10 distinct members in Group A and 6 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 7 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(16,7)=11,440 unrestricted committees. The all-A count is C(10,7)=120; the all-B count is C(6,7)=0. Excluding both forbidden one-group categories gives 11,440−120−0=11,320 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(10,1)C(6,6)=10; C(10,2)C(6,5)=270; C(10,3)C(6,4)=1,800; C(10,4)C(6,3)=4,200; C(10,5)C(6,2)=3,780; C(10,6)C(6,1)=1,260. Adding the contributions gives 11,320, matching the complement result. For comparison, “at least one from A” alone would allow 11,440 committees, while “exactly one from A” would allow 10. If k exceeds both group sizes, all possible committees automatically include both groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

9. Group A has 11, Group B has 3, choose 4

Question. A fictional activity has 11 distinct members in Group A and 3 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 4 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(14,4)=1,001 unrestricted committees. The all-A count is C(11,4)=330; the all-B count is C(3,4)=0. Excluding both forbidden one-group categories gives 1,001−330−0=671 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(11,1)C(3,3)=11; C(11,2)C(3,2)=165; C(11,3)C(3,1)=495. Adding the contributions gives 671, matching the complement result. For comparison, “at least one from A” alone would allow 1,001 committees, while “exactly one from A” would allow 11. Compare complement counting with a direct case split before trusting a long factorial calculation. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

10. Group A has 3, Group B has 10, choose 9

Question. A fictional activity has 3 distinct members in Group A and 10 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 9 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(13,9)=715 unrestricted committees. The all-A count is C(3,9)=0; the all-B count is C(10,9)=10. Excluding both forbidden one-group categories gives 715−0−10=705 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(3,1)C(10,8)=135; C(3,2)C(10,7)=360; C(3,3)C(10,6)=210. Adding the contributions gives 705, matching the complement result. For comparison, “at least one from A” alone would allow 705 committees, while “exactly one from A” would allow 135. If the groups overlap, this two-disjoint-group formula cannot be applied without reclassifying people. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

11. Group A has 4, Group B has 7, choose 6

Question. A fictional activity has 4 distinct members in Group A and 7 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(11,6)=462 unrestricted committees. The all-A count is C(4,6)=0; the all-B count is C(7,6)=7. Excluding both forbidden one-group categories gives 462−0−7=455 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(4,1)C(7,5)=84; C(4,2)C(7,4)=210; C(4,3)C(7,3)=140; C(4,4)C(7,2)=21. Adding the contributions gives 455, matching the complement result. For comparison, “at least one from A” alone would allow 455 committees, while “exactly one from A” would allow 84. Use combinations because the internal order of a committee does not matter. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

12. Group A has 5, Group B has 4, choose 6

Question. A fictional activity has 5 distinct members in Group A and 4 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(9,6)=84 unrestricted committees. The all-A count is C(5,6)=0; the all-B count is C(4,6)=0. Excluding both forbidden one-group categories gives 84−0−0=84 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(5,2)C(4,4)=10; C(5,3)C(4,3)=40; C(5,4)C(4,2)=30; C(5,5)C(4,1)=4. Adding the contributions gives 84, matching the complement result. For comparison, “at least one from A” alone would allow 84 committees, while “exactly one from A” would allow 0. The all-A and all-B excluded categories are disjoint for a nonempty committee drawn from disjoint groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

13. Group A has 6, Group B has 11, choose 8

Question. A fictional activity has 6 distinct members in Group A and 11 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 8 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(17,8)=24,310 unrestricted committees. The all-A count is C(6,8)=0; the all-B count is C(11,8)=165. Excluding both forbidden one-group categories gives 24,310−0−165=24,145 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(6,1)C(11,7)=1,980; C(6,2)C(11,6)=6,930; C(6,3)C(11,5)=9,240; C(6,4)C(11,4)=4,950; C(6,5)C(11,3)=990; C(6,6)C(11,2)=55. Adding the contributions gives 24,145, matching the complement result. For comparison, “at least one from A” alone would allow 24,145 committees, while “exactly one from A” would allow 1,980. When k exceeds a group’s size, selecting k entirely from that group contributes zero. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

14. Group A has 7, Group B has 8, choose 5

Question. A fictional activity has 7 distinct members in Group A and 8 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 5 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(15,5)=3,003 unrestricted committees. The all-A count is C(7,5)=21; the all-B count is C(8,5)=56. Excluding both forbidden one-group categories gives 3,003−21−56=2,926 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(7,1)C(8,4)=490; C(7,2)C(8,3)=1,176; C(7,3)C(8,2)=980; C(7,4)C(8,1)=280. Adding the contributions gives 2,926, matching the complement result. For comparison, “at least one from A” alone would allow 2,947 committees, while “exactly one from A” would allow 490. The direct split method counts j from A and k−j from B, with j restricted to feasible integers. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

15. Group A has 8, Group B has 5, choose 2

Question. A fictional activity has 8 distinct members in Group A and 5 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 2 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(13,2)=78 unrestricted committees. The all-A count is C(8,2)=28; the all-B count is C(5,2)=10. Excluding both forbidden one-group categories gives 78−28−10=40 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(8,1)C(5,1)=40. Adding the contributions gives 40, matching the complement result. For comparison, “at least one from A” alone would allow 68 committees, while “exactly one from A” would allow 40. A numerical answer larger than the unrestricted total signals a counting error. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

16. Group A has 9, Group B has 2, choose 7

Question. A fictional activity has 9 distinct members in Group A and 2 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 7 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(11,7)=330 unrestricted committees. The all-A count is C(9,7)=36; the all-B count is C(2,7)=0. Excluding both forbidden one-group categories gives 330−36−0=294 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(9,5)C(2,2)=126; C(9,6)C(2,1)=168. Adding the contributions gives 294, matching the complement result. For comparison, “at least one from A” alone would allow 330 committees, while “exactly one from A” would allow 0. Changing “at least one from each” into “exactly one from A” changes the permitted split cases. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

17. Group A has 10, Group B has 9, choose 4

Question. A fictional activity has 10 distinct members in Group A and 9 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 4 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(19,4)=3,876 unrestricted committees. The all-A count is C(10,4)=210; the all-B count is C(9,4)=126. Excluding both forbidden one-group categories gives 3,876−210−126=3,540 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(10,1)C(9,3)=840; C(10,2)C(9,2)=1,620; C(10,3)C(9,1)=1,080. Adding the contributions gives 3,540, matching the complement result. For comparison, “at least one from A” alone would allow 3,750 committees, while “exactly one from A” would allow 840. No extra k! factor belongs in an unranked committee selection. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

18. Group A has 11, Group B has 6, choose 9

Question. A fictional activity has 11 distinct members in Group A and 6 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 9 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(17,9)=24,310 unrestricted committees. The all-A count is C(11,9)=55; the all-B count is C(6,9)=0. Excluding both forbidden one-group categories gives 24,310−55−0=24,255 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(11,3)C(6,6)=165; C(11,4)C(6,5)=1,980; C(11,5)C(6,4)=6,930; C(11,6)C(6,3)=9,240; C(11,7)C(6,2)=4,950; C(11,8)C(6,1)=990. Adding the contributions gives 24,255, matching the complement result. For comparison, “at least one from A” alone would allow 24,310 committees, while “exactly one from A” would allow 0. If k exceeds both group sizes, all possible committees automatically include both groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

19. Group A has 3, Group B has 3, choose 4

Question. A fictional activity has 3 distinct members in Group A and 3 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 4 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(6,4)=15 unrestricted committees. The all-A count is C(3,4)=0; the all-B count is C(3,4)=0. Excluding both forbidden one-group categories gives 15−0−0=15 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(3,1)C(3,3)=3; C(3,2)C(3,2)=9; C(3,3)C(3,1)=3. Adding the contributions gives 15, matching the complement result. For comparison, “at least one from A” alone would allow 15 committees, while “exactly one from A” would allow 3. Compare complement counting with a direct case split before trusting a long factorial calculation. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

20. Group A has 4, Group B has 10, choose 3

Question. A fictional activity has 4 distinct members in Group A and 10 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 3 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(14,3)=364 unrestricted committees. The all-A count is C(4,3)=4; the all-B count is C(10,3)=120. Excluding both forbidden one-group categories gives 364−4−120=240 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(4,1)C(10,2)=180; C(4,2)C(10,1)=60. Adding the contributions gives 240, matching the complement result. For comparison, “at least one from A” alone would allow 244 committees, while “exactly one from A” would allow 180. If the groups overlap, this two-disjoint-group formula cannot be applied without reclassifying people. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

21. Group A has 5, Group B has 7, choose 8

Question. A fictional activity has 5 distinct members in Group A and 7 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 8 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(12,8)=495 unrestricted committees. The all-A count is C(5,8)=0; the all-B count is C(7,8)=0. Excluding both forbidden one-group categories gives 495−0−0=495 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(5,1)C(7,7)=5; C(5,2)C(7,6)=70; C(5,3)C(7,5)=210; C(5,4)C(7,4)=175; C(5,5)C(7,3)=35. Adding the contributions gives 495, matching the complement result. For comparison, “at least one from A” alone would allow 495 committees, while “exactly one from A” would allow 5. Use combinations because the internal order of a committee does not matter. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

22. Group A has 6, Group B has 4, choose 5

Question. A fictional activity has 6 distinct members in Group A and 4 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 5 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(10,5)=252 unrestricted committees. The all-A count is C(6,5)=6; the all-B count is C(4,5)=0. Excluding both forbidden one-group categories gives 252−6−0=246 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(6,1)C(4,4)=6; C(6,2)C(4,3)=60; C(6,3)C(4,2)=120; C(6,4)C(4,1)=60. Adding the contributions gives 246, matching the complement result. For comparison, “at least one from A” alone would allow 252 committees, while “exactly one from A” would allow 6. The all-A and all-B excluded categories are disjoint for a nonempty committee drawn from disjoint groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

23. Group A has 7, Group B has 11, choose 12

Question. A fictional activity has 7 distinct members in Group A and 11 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 12 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(18,12)=18,564 unrestricted committees. The all-A count is C(7,12)=0; the all-B count is C(11,12)=0. Excluding both forbidden one-group categories gives 18,564−0−0=18,564 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(7,1)C(11,11)=7; C(7,2)C(11,10)=231; C(7,3)C(11,9)=1,925; C(7,4)C(11,8)=5,775; C(7,5)C(11,7)=6,930; C(7,6)C(11,6)=3,234; C(7,7)C(11,5)=462. Adding the contributions gives 18,564, matching the complement result. For comparison, “at least one from A” alone would allow 18,564 committees, while “exactly one from A” would allow 7. When k exceeds a group’s size, selecting k entirely from that group contributes zero. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

24. Group A has 8, Group B has 8, choose 7

Question. A fictional activity has 8 distinct members in Group A and 8 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 7 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(16,7)=11,440 unrestricted committees. The all-A count is C(8,7)=8; the all-B count is C(8,7)=8. Excluding both forbidden one-group categories gives 11,440−8−8=11,424 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(8,1)C(8,6)=224; C(8,2)C(8,5)=1,568; C(8,3)C(8,4)=3,920; C(8,4)C(8,3)=3,920; C(8,5)C(8,2)=1,568; C(8,6)C(8,1)=224. Adding the contributions gives 11,424, matching the complement result. For comparison, “at least one from A” alone would allow 11,432 committees, while “exactly one from A” would allow 224. The direct split method counts j from A and k−j from B, with j restricted to feasible integers. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

25. Group A has 9, Group B has 5, choose 4

Question. A fictional activity has 9 distinct members in Group A and 5 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 4 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(14,4)=1,001 unrestricted committees. The all-A count is C(9,4)=126; the all-B count is C(5,4)=5. Excluding both forbidden one-group categories gives 1,001−126−5=870 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(9,1)C(5,3)=90; C(9,2)C(5,2)=360; C(9,3)C(5,1)=420. Adding the contributions gives 870, matching the complement result. For comparison, “at least one from A” alone would allow 996 committees, while “exactly one from A” would allow 90. A numerical answer larger than the unrestricted total signals a counting error. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

26. Group A has 10, Group B has 2, choose 9

Question. A fictional activity has 10 distinct members in Group A and 2 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 9 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(12,9)=220 unrestricted committees. The all-A count is C(10,9)=10; the all-B count is C(2,9)=0. Excluding both forbidden one-group categories gives 220−10−0=210 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(10,7)C(2,2)=120; C(10,8)C(2,1)=90. Adding the contributions gives 210, matching the complement result. For comparison, “at least one from A” alone would allow 220 committees, while “exactly one from A” would allow 0. Changing “at least one from each” into “exactly one from A” changes the permitted split cases. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

27. Group A has 11, Group B has 9, choose 6

Question. A fictional activity has 11 distinct members in Group A and 9 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(20,6)=38,760 unrestricted committees. The all-A count is C(11,6)=462; the all-B count is C(9,6)=84. Excluding both forbidden one-group categories gives 38,760−462−84=38,214 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(11,1)C(9,5)=1,386; C(11,2)C(9,4)=6,930; C(11,3)C(9,3)=13,860; C(11,4)C(9,2)=11,880; C(11,5)C(9,1)=4,158. Adding the contributions gives 38,214, matching the complement result. For comparison, “at least one from A” alone would allow 38,676 committees, while “exactly one from A” would allow 1,386. No extra k! factor belongs in an unranked committee selection. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

28. Group A has 3, Group B has 6, choose 3

Question. A fictional activity has 3 distinct members in Group A and 6 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 3 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(9,3)=84 unrestricted committees. The all-A count is C(3,3)=1; the all-B count is C(6,3)=20. Excluding both forbidden one-group categories gives 84−1−20=63 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(3,1)C(6,2)=45; C(3,2)C(6,1)=18. Adding the contributions gives 63, matching the complement result. For comparison, “at least one from A” alone would allow 64 committees, while “exactly one from A” would allow 45. If k exceeds both group sizes, all possible committees automatically include both groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

29. Group A has 4, Group B has 3, choose 6

Question. A fictional activity has 4 distinct members in Group A and 3 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(7,6)=7 unrestricted committees. The all-A count is C(4,6)=0; the all-B count is C(3,6)=0. Excluding both forbidden one-group categories gives 7−0−0=7 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(4,3)C(3,3)=4; C(4,4)C(3,2)=3. Adding the contributions gives 7, matching the complement result. For comparison, “at least one from A” alone would allow 7 committees, while “exactly one from A” would allow 0. Compare complement counting with a direct case split before trusting a long factorial calculation. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

30. Group A has 5, Group B has 10, choose 5

Question. A fictional activity has 5 distinct members in Group A and 10 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 5 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(15,5)=3,003 unrestricted committees. The all-A count is C(5,5)=1; the all-B count is C(10,5)=252. Excluding both forbidden one-group categories gives 3,003−1−252=2,750 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(5,1)C(10,4)=1,050; C(5,2)C(10,3)=1,200; C(5,3)C(10,2)=450; C(5,4)C(10,1)=50. Adding the contributions gives 2,750, matching the complement result. For comparison, “at least one from A” alone would allow 2,751 committees, while “exactly one from A” would allow 1,050. If the groups overlap, this two-disjoint-group formula cannot be applied without reclassifying people. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

31. Group A has 6, Group B has 7, choose 2

Question. A fictional activity has 6 distinct members in Group A and 7 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 2 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(13,2)=78 unrestricted committees. The all-A count is C(6,2)=15; the all-B count is C(7,2)=21. Excluding both forbidden one-group categories gives 78−15−21=42 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(6,1)C(7,1)=42. Adding the contributions gives 42, matching the complement result. For comparison, “at least one from A” alone would allow 57 committees, while “exactly one from A” would allow 42. Use combinations because the internal order of a committee does not matter. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

32. Group A has 7, Group B has 4, choose 7

Question. A fictional activity has 7 distinct members in Group A and 4 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 7 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(11,7)=330 unrestricted committees. The all-A count is C(7,7)=1; the all-B count is C(4,7)=0. Excluding both forbidden one-group categories gives 330−1−0=329 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(7,3)C(4,4)=35; C(7,4)C(4,3)=140; C(7,5)C(4,2)=126; C(7,6)C(4,1)=28. Adding the contributions gives 329, matching the complement result. For comparison, “at least one from A” alone would allow 330 committees, while “exactly one from A” would allow 0. The all-A and all-B excluded categories are disjoint for a nonempty committee drawn from disjoint groups. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

33. Group A has 8, Group B has 11, choose 4

Question. A fictional activity has 8 distinct members in Group A and 11 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 4 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(19,4)=3,876 unrestricted committees. The all-A count is C(8,4)=70; the all-B count is C(11,4)=330. Excluding both forbidden one-group categories gives 3,876−70−330=3,476 valid committees. Both forbidden categories are feasible but disjoint, so the subtraction does not require an overlap correction.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(8,1)C(11,3)=1,320; C(8,2)C(11,2)=1,540; C(8,3)C(11,1)=616. Adding the contributions gives 3,476, matching the complement result. For comparison, “at least one from A” alone would allow 3,546 committees, while “exactly one from A” would allow 1,320. When k exceeds a group’s size, selecting k entirely from that group contributes zero. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

34. Group A has 9, Group B has 8, choose 10

Question. A fictional activity has 9 distinct members in Group A and 8 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 10 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(17,10)=19,448 unrestricted committees. The all-A count is C(9,10)=0; the all-B count is C(8,10)=0. Excluding both forbidden one-group categories gives 19,448−0−0=19,448 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(9,2)C(8,8)=36; C(9,3)C(8,7)=672; C(9,4)C(8,6)=3,528; C(9,5)C(8,5)=7,056; C(9,6)C(8,4)=5,880; C(9,7)C(8,3)=2,016; C(9,8)C(8,2)=252; C(9,9)C(8,1)=8. Adding the contributions gives 19,448, matching the complement result. For comparison, “at least one from A” alone would allow 19,448 committees, while “exactly one from A” would allow 0. The direct split method counts j from A and k−j from B, with j restricted to feasible integers. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

35. Group A has 10, Group B has 5, choose 6

Question. A fictional activity has 10 distinct members in Group A and 5 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 6 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(15,6)=5,005 unrestricted committees. The all-A count is C(10,6)=210; the all-B count is C(5,6)=0. Excluding both forbidden one-group categories gives 5,005−210−0=4,795 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(10,1)C(5,5)=10; C(10,2)C(5,4)=225; C(10,3)C(5,3)=1,200; C(10,4)C(5,2)=2,100; C(10,5)C(5,1)=1,260. Adding the contributions gives 4,795, matching the complement result. For comparison, “at least one from A” alone would allow 5,005 committees, while “exactly one from A” would allow 10. A numerical answer larger than the unrestricted total signals a counting error. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

36. Group A has 11, Group B has 2, choose 3

Question. A fictional activity has 11 distinct members in Group A and 2 distinct members in Group B. The groups do not overlap. Select an unranked committee of exactly 3 people containing at least one member from each group. Calculate the number of possible committees by complement counting, then verify by splitting into exact group counts. Do not introduce a seating order or imaginary officer roles.

Complement calculation. There are C(13,3)=286 unrestricted committees. The all-A count is C(11,3)=165; the all-B count is C(2,3)=0. Excluding both forbidden one-group categories gives 286−165−0=121 valid committees. At least one forbidden category is impossible because that group contains too few members for an all-one-group committee.

Direct independent verification. Classify every valid committee by the exact number j selected from A. The feasible counts are C(11,1)C(2,2)=11; C(11,2)C(2,1)=110. Adding the contributions gives 121, matching the complement result. For comparison, “at least one from A” alone would allow 286 committees, while “exactly one from A” would allow 11. Changing “at least one from each” into “exactly one from A” changes the permitted split cases. Changing the group sizes or committee size changes which j values are legal; redo that constraint check before reusing the numerical result.

Ten quick answers for revision

Five from A, four from B, choose three: 84−10−4=70.
In that example, one A plus two B contributes 30 committees.
Two A plus one B contributes the remaining 40.
Six A and four B, choose five: 252−6−0=246.
Four A and two B, choose four: 15−1−0=14.
Three A and three B, choose two: C(3,1)C(3,1)=9.
Six A and six B, choose seven: all 792 committees contain both groups.
With five A and four B, choose three, “at least one from A” gives 80.
With five A and four B, choose three, “exactly one from A” gives 30.
The 2027 H2 Mathematics syllabus includes combinations and selection restrictions.

Five-day revision and three-learner practice

Day one: distinguish an unranked committee from an ordered assignment of officers, using simple C(n,k) examples. Day two: learn the two-group complement formula and identify the all-A and all-B forbidden categories. Day three: verify answers using direct exact-j splits, checking feasibility of each group count. Day four: compare “at least one A”, “at least one each” and “exactly one A” problems. Day five: include boundary conditions where a group is too small to fill the whole committee, then explain why its impossible selection count is zero.

In a three-learner tutorial, one pupil can count all unrestricted committees, another identify and subtract the forbidden categories, and a third verify with a direct group-split sum. Rotate roles. The goal is to reason from the event description rather than memorise C(a+b,k)−C(a,k)−C(b,k) without knowing when it applies.

Frequently asked questions

Why do committees use combinations rather than permutations?

An ordinary committee is a set of members without internal order. If named roles matter, the finished counting object is different.

Why subtract all-A and all-B committees?

Those are exactly the committees missing one of the two groups, so removing them leaves committees with at least one from each.

Can I subtract both categories without adding an overlap?

Yes for a nonempty committee and two disjoint groups: a committee cannot be entirely from A and entirely from B at once.

What if k is larger than one group?

The number of all-from-that-group selections is zero. Use C(n,k)=0 for k>n.

How do I count directly?

Sum C(a,j)C(b,k−j) over all integers j with at least one selected from each group and enough people available.

Does “at least one A” mean the same as “at least one from each”?

No. The first allows all-A committees; the second requires representation of B as well.

What if a person belongs to both groups?

Reclassify the overlapping membership before counting. The simple disjoint-group formula would not apply unchanged.

Is this part of 2027 H2 Mathematics?

Yes. Section 6.1 includes permutation and combination techniques and counting arrangements subject to restrictions.

JC2 H2 Mathematics Tutor Bukit Timah | Committee Selection: At Least One From Each Group