eduKate Singapore

VIEW THIS AS

Auto mode follows the Route Engine until you choose a viewpoint.

YOU ARE HERE

ROUTE CHECK

CONNECTED TO

WHAT NEXT

Use the canonical route for this room, or HELP if you are unsure.

Next useful routeCanonical hubHELP

Mathematics in Bukit Timah | Counting Grid Paths with Pascal’s Triangle and Combinations

—

by

eduKate Asia
Original conceptual illustration of a person following a self-guided nature audio walk with earphones and a printed trail map in a tropical botanical garden; not an actual site photograph.

Suppose a fictional neighbourhood map has straight east–west and north–south paths arranged as a grid. A student begins at the southwest corner and must reach the northeast corner, moving only east or north. How many different shortest routes are possible? The answer depends not on how many roads appear attractive, but on how many orders of the required steps are allowed.

This Mathematics in Bukit Timah lesson uses an imagined rectangular grid, not a survey of actual roads, walkways, park entrances or accessible routes in Bukit Timah. Places such as Sixth Avenue and King Albert Park provide a familiar setting for thinking about directions, but every junction, route length, restriction and travel time in the problems below is invented for teaching. Never interpret the grid as a real navigation map.

The quick answer: if a shortest route requires two east steps (E) and two north steps (N), there are six distinct orders: EENN, ENEN, ENNE, NEEN, NENE and NNEE. For larger grids, combinations and Pascal’s triangle allow students to count efficiently without listing every route.

Article contents

Original conceptual illustration of a person with a printed route map in a tropical garden; the rectangular street grid and grid-path counts in this article are entirely fictional.

Start with the restrictions, not the formula

A route-counting question must specify where the journey starts and ends, which moves are permitted and whether different orders count as different routes. Without those rules, “How many paths?” may have no finite answer.

In this model, east increases the horizontal grid coordinate by one, north increases the vertical coordinate by one, and a shortest journey never goes west or south. Two journeys differ when their sequences of E and N moves differ.

The phrase “shortest” is crucial. Allowing loops or return steps introduces many additional journeys and may even permit infinitely many walk sequences in a graph where revisiting is allowed. The simple binomial count does not describe that broader task.

Why use Bukit Timah as a mathematical story?

The local setting is a way to make coordinate movement imaginable. A fictional grid might be described as an exercise based near the wider Bukit Timah area, but the street network there is not a uniform square lattice.

No actual street intersections, walking access points or journey distances are inferred from this article. To plan a real journey, use an appropriate current map and local access guidance instead of the invented grid.

The value for pupils is learning to abstract one mathematical structure from a story. Once they understand restricted moves on graph paper, they can solve a new counting problem without needing to know the neighbourhood at all.

Worked example 1: one east and one north

Start at (0,0) and finish at (1,1) on an imaginary square grid. Two moves are required: exactly one E and exactly one N.

The possible orders are EN and NE. Both arrive at the same endpoint, but the route traces different intermediate junctions. Therefore the answer is two.

This is a manageable first exercise for learners who are not yet ready for factorial notation. Paper arrows show why order matters, and students can check each sequence against a drawing.

Two east moves and one north move

Now move from (0,0) to (2,1). Every shortest path needs two E moves and one N move, for three moves in total.

The distinct sequences are EEN, ENE and NEE. Writing “EEN” twice with the two east steps relabelled as E₁ and E₂ does not create new physical routes: those east moves are identical in the model.

The answer is three, not six. This reveals why simply calculating 3! overcounts arrangements when repeated step types are indistinguishable.

Worked example 2: arrange two of each step

From (0,0) to (2,2), the learner must perform two east and two north moves. There are four positions in the sequence, and the question is which two positions are occupied by E.

Choose any two of the four positions for E; the remaining two positions must be N. This gives 4 choose 2 = 6. Listing the sequences confirms all six possibilities.

That second method is a bridge from enumeration to combinations. It counts the placements of identical moves without assigning artificial identities to those moves.

The binomial coefficient counts selected positions

The notation \(\binom{n}{r}\) is read “n choose r”. It counts ways to choose r positions from n distinct positions without considering an order among the selected positions.

For a shortest grid path consisting of e east steps and n north steps, total moves equal e+n. Choosing the positions of the east steps gives \(\binom{e+n}{e}\); choosing north positions instead gives \(\binom{e+n}{n}\).

These expressions are equal because fixing the positions of one type automatically fixes those of the other. The formula is a compressed statement of the counting logic, not something to use before understanding the route restrictions.

Worked example 3: three east and two north

An imaginary start at (0,0) and finish at (3,2) requires five moves: E,E,E,N,N.

Select two of the five sequence positions for the north moves. The number of shortest routes is \(\binom{5}{2}=5!/(2!3!)=10\).

A child can verify a few examples—EENEN, NEEEN, ENENE—and check that each contains exactly three east moves and two north moves. The factorial formula allows the ten routes to be counted without exhaustively listing them.

Why factorials must be divided for repeated moves

If five moves were all distinctly labelled, there would be 5! = 120 arrangements. But our three E moves are identical, and our two N moves are identical.

Permuting the three east labels leaves a route unchanged, producing 3! duplicate labellings for each physical order. The two north labels produce another 2! duplicate factor. Divide 120 by 3!×2! = 12 to obtain ten.

This explanation prevents the frequent error of reporting 120 simply because five moves appear in a question.

Counting shortest paths with a table

A rectangular lattice offers a method that avoids factorial calculations. Write the number of ways to reach each junction from the southwest starting point.

Every non-start junction can be reached from the junction immediately west or immediately south, provided those junctions and incoming moves are allowed. Add their counts to obtain the current count.

At the first row or column, where only one direction is possible, counts are one. Continuing this addition fills a grid of route counts and reproduces the values of Pascal’s triangle.

Worked example 4: the two-by-two counting grid

Begin by writing 1 at the start. Along the south and west boundaries, every reachable junction has count 1 because only a straight sequence of identical steps gets there.

The interior junction at (1,1) has 2 routes: one through (0,1) and one through (1,0). Junction (2,1) receives 2+1=3; junction (1,2) also receives 1+2=3.

The final (2,2) junction has 3+3=6 incoming paths. The table agrees with the six listed sequences. A picture can therefore explain a combination without writing any factorial.

Pascal’s triangle emerges from adding neighbours

Pascal’s triangle starts with rows such as 1; 1,1; 1,2,1; 1,3,3,1; and 1,4,6,4,1.

Each interior number equals the two numbers just above it added together. The route-counting table uses the same logic: to arrive at a junction, a valid last move must come from one of two predecessor junctions.

The coincidence is structural, not a trick with familiar number patterns. Both situations classify outcomes by their last or selected position and combine disjoint possibilities.

Worked example 5: a four-move route count

If a shortest journey has one east and three north steps, it has four positions for its one E. That step can occur first, second, third or fourth.

There are therefore four sequences, matching \(\binom{4}{1}=4\) and the Pascal’s triangle row 1,4,6,4,1.

If the student writes “four” but cannot explain where the E can be placed, ask for the four sequences. This small case confirms the meaning of the formula before more complicated questions begin.

The symmetry of combinations has a route meaning

Why is \(\binom{5}{2}=\binom{5}{3}\)? In a five-step shortest path with two north and three east moves, selecting the two north positions leaves exactly three east positions.

Selecting the three east positions instead identifies the same routes, simply from the complementary side of the decision. Neither count should be larger.

This relationship is a useful error check. If a student calculates two different values from the two expressions, at least one calculation is wrong.

Shortest by steps is not necessarily fastest by minutes

A fictional rectangular grid can be constructed so that every east and north edge is the same length. Under that assumption, all monotone routes between fixed endpoints use the same number of steps and cover the same distance.

Real journeys may differ in waiting, gradients, crossings, closures or speed. The number of shortest grid-step routes does not identify the fastest actual journey.

This distinction links to Mathematics in Bukit Timah: distance, time and speed. One article models numerical rates; this one counts the arrangements of allowed moves.

A blocked junction changes the counting problem

Suppose the fictional route from (0,0) to (3,2) would normally have ten shortest paths. Now imagine that junction (1,1) is closed and cannot be visited.

The routes through (1,1) can be counted as paths from start to (1,1) multiplied by paths from (1,1) to finish. The first part has \(\binom{2}{1}=2\) paths; the second needs two E and one N, giving \(\binom{3}{1}=3\).

Thus 2×3=6 shortest paths pass through the closed junction, and 10−6=4 avoid it. The junction and closure are invented mathematical conditions, not real Bukit Timah access information.

Worked example 6: why multiplication applies at a required stop

When a route is required to pass through a particular junction, it has two stages: start to junction, then junction to destination.

If there are two valid ways to complete the first stage and three valid ways to complete the second, and every first-stage choice can be paired with every second-stage choice, the multiplication principle gives 2×3=6 routes.

The independence here is combinatorial compatibility, not necessarily statistical independence. Check that stages fit together and that no restrictions make some combinations invalid before multiplying their counts.

Subtraction works only when the excluded routes are counted correctly

Start with all ten shortest paths from (0,0) to (3,2). Remove the six that pass through (1,1). The four remaining paths avoid the forbidden junction.

A student who merely subtracts one for the blocked junction has confused a location with a route. A single location may lie on many different paths.

The relevant counting objects must stay clear throughout: junctions are points; routes are complete sequences of edges. Excluding one point can exclude several complete sequences.

Two forbidden points may require inclusion–exclusion

If a question bans two junctions A and B, counting routes through A and through B separately and subtracting both from the total may remove routes that pass through both junctions twice.

A corrected count of routes avoiding both is total minus routes through A minus routes through B plus routes through both, where those sets are counted under compatible movement rules.

This is the same inclusion–exclusion principle used in Bukit Timah Venn diagrams, now applied to sets of routes instead of sets of pupils.

Worked example 7: two aligned checkpoints

Imagine a grid trip from (0,0) to (4,3), with required checkpoints A=(1,1) and B=(3,2). Every allowed move remains east or north.

From start to A there are \(\binom{2}{1}=2\) shortest paths. From A to B there are two E and one N, giving \(\binom{3}{1}=3\). From B to end there are one E and one N, giving 2 paths.

The number of paths that visit A then B is 2×3×2=12. The checkpoints have compatible increasing coordinates, so all these path segments can be joined without backtracking.

A checkpoint in the wrong direction can make the count zero

Suppose a route moving only east or north must visit A=(3,2) and then B=(1,3). From A to B, the x-coordinate would have to decrease from 3 to 1.

That requires westward movement, which is forbidden. Therefore no valid monotone route can visit the checkpoints in that order.

This is a logic check before a formula check. If a student’s factorial calculation gives a positive number, it has ignored the movement restrictions.

A permutation usually describes arranging distinct objects, while a grid path here arranges repeated move types. An ordinary n! calculation counts permutations of n distinct labels.

A binomial coefficient counts which of the n positions receive one of two identical move types. It is often described as a combination because selecting positions is an unordered choice.

The key is to define what counts as a distinct outcome. If there are actual different tasks or named checkpoints, the appropriate permutation or combination model may change.

A restriction on the first step reduces the count

Consider the fictional (0,0) to (3,2) trip with ten routes. Suppose the first move must be east. After that first E, there remain two E and two N moves.

Their arrangements number \(\binom{4}{2}=6\). Therefore six of the original ten routes begin east. The remaining four begin north.

The two counts add to ten because every valid shortest route begins either east or north and no route can have both first moves.

Worked example 8: exactly one chosen first direction

What is the number of routes from (0,0) to (3,2) beginning with north? After the first N, there remain three E and one N.

Arrange the remaining four moves by selecting the position of the remaining N: \(\binom{4}{1}=4\). This agrees with total ten minus the six east-first routes.

The two-method check is valuable. It tests whether the condition was correctly translated into a fixed initial move.

The probability of a route needs a selection rule

There are six east-first and four north-first paths in the fictional five-step model. If all ten complete routes are chosen uniformly, the probability of an east-first route is 6/10.

But if a walker chooses east or north with equal probability at each eligible junction, the resulting distribution over complete routes may not be uniform. Some sequences can be more likely than others.

Counting the routes does not automatically establish probabilities. This is a distinct connection to the Bukit Timah probability and fair games guide.

Worked example 9: uniform complete-path selection

Suppose ten hypothetical monotone shortest paths exist and a computer is explicitly programmed to choose each one with equal probability.

Six begin east, so P(first east)=6/10=3/5. Four begin north, so P(first north)=4/10=2/5. Their probabilities sum to one.

Without that uniform-selection assumption, those ratios are not guaranteed. The distinction is important because a correct counting denominator can still be the wrong probability denominator.

The number of routes grows faster than the grid looks

For a square grid requiring n east and n north moves, the number of monotone shortest routes is \(\binom{2n}{n}\).

For n=1 the count is 2; n=2 gives 6; n=3 gives 20; n=4 gives 70; and n=5 gives 252. The increase is rapid even though each grid is only one unit larger in both directions.

This makes a good estimating exercise. A student can recognise that manually listing every route becomes impractical well before the diagram becomes visually complicated.

Worked example 10: four east and four north

A trip requiring four E and four N has eight moves. Select four of the eight positions for E.

The number of routes is \(\binom{8}{4}=8!/(4!4!)=70\). It is not 8!=40,320, because moves of the same type are repeated and indistinguishable.

A complete enumeration would be time-consuming; the formula gives the count directly once the underlying choice has been justified.

Dynamic counting can accommodate unequal local restrictions

Suppose a fictional grid has several forbidden edges or inaccessible junctions. The simple closed combination formula may no longer apply.

A counting table can still work if each junction count is the sum of counts from allowed predecessor junctions. A forbidden junction gets zero; a missing edge contributes no incoming routes from that direction.

This dynamic-programming idea is accessible through repeated addition. Students do not need computer code to understand the logic, although a spreadsheet can help with larger grids.

Worked example 11: a small grid with one closed edge

Consider the (0,0) to (2,2) grid. Normally there are six shortest routes. Ban only the directed edge from (1,0) to (1,1), while keeping all junctions open.

To use that edge, a path must first reach (1,0), which has one shortest prefix E; then traverse the banned north step; then travel from (1,1) to (2,2), which has two completions EN or NE.

Exactly two routes use the banned edge, so four remain. Blocking a single edge is different from blocking the entire junction, which would remove more possibilities.

Do not confuse a route with its reversal

In the model, the allowed moves are north and east, so the reverse of a valid route would travel south and west and would not satisfy the same start-to-end restrictions.

If a different problem allows undirected movement and counts travel between two endpoints without distinguishing direction, it must define whether a route and its reversal are regarded as the same object.

Counting conventions have consequences. Students should not silently divide an answer by two because a path can be traversed in reverse under some unrelated interpretation.

A shortest-step route and a shortest-distance route require a scale

In a uniform grid, one E or N step may be defined as the same hypothetical distance, for instance 100 metres. Then every route requiring three E and two N covers five steps, or 500 metres, along the grid edges.

If horizontal and vertical grid intervals differ in length, every monotone route still uses the same count of E and N, so its total edge distance is identical given fixed east and north segment lengths. But a direct straight line between endpoints is shorter than the grid-edge walk when both components are nonzero.

The geometric distance formula is explored separately in this series; route counting does not need to pretend the grid is an actual street network.

Why diagonal movement changes the answer

If diagonal northeast moves are permitted, a journey to (2,2) can use fewer than four steps. Two diagonal moves would reach the destination in two steps if each diagonal move counts as one.

The shortest-step paths under the new rules are no longer the six E/N sequences. Their count and distance depend on how diagonal edges are defined and weighted.

Before counting, inspect the movement alphabet. Adding one permitted move type can change both the minimum journey length and the set of shortest routes.

A grid has a beginning and an ending, not just a shape

Two grids can have identical dimensions but different starting corners or destinations. The required east and north step totals are determined by the displacement between the specified endpoints.

If start is (1,1) and destination is (4,3), the displacement is three E and two N, so ten monotone shortest routes exist. The location of the whole drawn grid’s external boundary is irrelevant if all required intermediate junctions are available.

This builds the coordinate skill of subtracting starting coordinates from finishing coordinates before counting.

A staircase path can be represented as a word

Sequences like ENEEN or NEEEN are mathematical words made from an alphabet of two symbols. Each valid sequence encodes one path when the start and grid rules are fixed.

This representation makes it easy to test a proposed answer. Count the E symbols and N symbols, then follow the moves. A route with an extra north step or a westward move belongs to a different problem.

It also allows pupils to explain why changing the order changes the intermediate route while keeping the endpoint unchanged.

A counting answer is not a recommendation for navigation

Even if a fictional grid has 70 shortest paths, an actual pedestrian network may have crossings, elevation changes, locked gates or safety restrictions. No claim about accessibility follows from a combinatorial count.

The article’s purpose is to teach constrained arrangement and exact enumeration. Practical travel needs independent, current information about the actual route.

Maintaining that distinction prevents a neat classroom result from being misapplied as real-world guidance.

Primary 2–4: learn by physically arranging arrows

Younger learners can begin with one east and one north arrow, then two east and one north. Move paper counters along drawn square grids and list distinct orders.

Ask whether changing two identical east arrows creates a new path; the correct answer is no. This builds the meaning of repetition before factorials or combinations are introduced.

A successful learner can show all two or three small-grid paths and explain why none has been duplicated.

Primary 5–6: tables and systematic listing

Older Primary students can organise routes by first move, check totals using a counting grid and investigate small entries of Pascal’s triangle as an enrichment activity suited to their prerequisites.

A useful task asks the child to separate the six paths to (2,2) into those beginning E and those beginning N. Each category has three routes and the disjoint categories cover the entire sample.

This approach develops systematic counting and clear reasoning without presenting advanced combinatorics as a required PSLE topic.

Secondary learners: from patterns to combinations

Secondary students with suitable prerequisites can connect route counts to binomial coefficients, factorials and the principle of multiplying compatible choices.

The intended assessment level should be checked against the applicable syllabus; combinations and factorial notation are not a universal requirement for every Secondary 1 class.

The deeper lesson is the transition from listing a small sample to proving a general formula. The equation should follow a clear statement of the objects being selected.

Upper-secondary extension: recurrence and Pascal identity

The identity \(\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}\) expresses a division into cases based on whether a chosen position is included.

In grid routes, classify each completed path by its final move. Routes ending east come from the predecessor to the west; routes ending north come from the predecessor to the south.

Because these cases cannot overlap and exhaust the allowed last steps, their counts add. That is the reason Pascal’s triangle works as a path-counting table.

Three errors tutors should diagnose separately

One child may list EENN and NEEN but miss other arrangements. Another may count relabelled identical east steps as different. A third may calculate the full-grid total correctly but forget that a forbidden junction invalidates several routes.

These are distinct errors of enumeration, indistinguishability and constraint handling. The correction should match the first invalid decision rather than simply provide the binomial formula.

A small-group tutorial can ask students to narrate one route, classify routes by the first step and then verify with an alternative counting method. Understanding becomes visible through the explanation.

A twenty-minute diagnostic for families

Minutes 1–5: From (0,0) to (1,1), list EN and NE. Ask why there are exactly two.

Minutes 6–10: From (0,0) to (2,1), list EEN, ENE and NEE. Ask why exchanging the names of two identical E steps changes nothing.

Minutes 11–15: From (0,0) to (2,2), count six shortest routes and fill the route-counting table using neighbour addition.

Minutes 16–20: Ban the central junction (1,1). Four of the six paths go through it, leaving two valid routes. Ask the child to justify this by counting prefixes and suffixes or checking the list.

A seven-day practice path with spaced retrieval

Day one focuses on defining valid moves. Day two lists small routes by hand. Day three groups routes by first move. Day four introduces the counting grid.

Day five connects counting tables to Pascal’s triangle. Day six adds a required checkpoint or one blocked junction. Day seven changes the grid size and asks for an independent explanation, not just a final number.

If the child cannot reliably track ordered steps, repair that prerequisite before introducing factorial notation.

Practice problems and checked answers

A. From (0,0) to (2,1), allowing only E and N, how many shortest paths? Answer: \(\binom{3}{1}=3\).

B. From (0,0) to (3,2), how many? Answer: \(\binom{5}{2}=10\).

C. From (0,0) to (3,2), first step fixed as E? Answer: \(\binom{4}{2}=6\).

D. From (0,0) to (2,2), avoiding the central junction (1,1)? Answer: 6−(2×2)=2.

E. From (0,0) to (4,4), how many unrestricted monotone shortest grid paths? Answer: \(\binom{8}{4}=70\).

Frequently asked questions

Does every shortest grid route have the same number of steps?

Under the specified east-and-north-only unit-grid model between fixed endpoints, yes. The endpoint displacement fixes how many moves of each type are needed.

Why do we not count all factorial permutations?

Because all east moves are identical and all north moves are identical. Labelling them separately would count the same actual sequence repeatedly.

What is Pascal’s triangle doing in a map problem?

Each junction can be reached from two predecessor directions, and the total count is the sum of the counts leading into it.

Is a blocked junction the same as a blocked edge?

No. A junction ban removes every route using that point, while an edge ban removes only routes that traverse the specified connection.

Do ten routes mean each has probability one-tenth?

Only if the experiment explicitly selects each complete route with equal probability.

Does this describe routes around Sixth Avenue MRT?

No. The grid and all junction restrictions are invented Mathematics examples, not a mapped walking network.

Is this part of every Primary Mathematics syllabus?

No. Younger children can use systematic listing and patterns; advanced combination notation depends on level and course.

What proves genuine understanding?

A learner can state the movement rules, explain what makes two routes different, count by two independent methods and adjust when a restriction changes.

Sources and next mathematical connections

For the underlying Singapore school progression, use the MOE Primary Mathematics syllabus and the SEAB 2027 G3 Mathematics syllabus directory. The advanced binomial work is an extension where suitable, not a claim about every student’s examination requirements.

Continue with Mathematics in Bukit Timah: compass bearings and route maps for interpreting directions, and Bukit Timah Venn diagrams for the inclusion–exclusion logic behind multiple forbidden junctions. The immutable Clementi Secondary 1 Mathematics small-group tutorial reference anchors the expectation that students learn why a method works before using it fluently.

The durable habit is to define the allowed moves, identify what counts as one distinct route, count systematically and test every restriction against the original model. That habit turns a simple imaginary map into genuine mathematical reasoning.

Continue this four-article Mathematics in Bukit Timah batch: Quadratic graphs, arches and turning points · Scatter plots, correlation and causation · Coordinate distance, midpoints and bisectors. These original guides use different mathematical models, with worked questions, learning diagnostics, careful assumptions and sources.

Discover more from eduKate Singapore

Subscribe now to keep reading and get access to the full archive.

Continue reading