BTT Mathematics / Primary Mathematics Learning Hub / Pascal’s Triangle and Combinations
Pascal’s Triangle is a compact counting machine. Each interior entry is the sum of the two entries above it. The same numbers appear when we count selections where order does not matter and when we count shortest grid paths made from two types of step.
In this guide, the top 1 is called Row 0. The next rows are:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
The convention matters because some books call the top row Row 1. Stating the convention prevents an indexing disagreement from becoming a mathematics error.
This guide is Primary Mathematics enrichment. It extends Systematic Listing and Logical Counting, Patterns and Early Algebra, and Optimisation and Minimum Moves. Combination notation such as C(n,r) or “n choose r” is optional bridge work; the counting ideas can be developed without formal notation.
Use the MOE Primary curriculum page and the learner’s school programme for required content.
Build the triangle · Read its patterns · Count selections · Count shortest paths · Why neighbouring counts add · 24 questions · Worked answers · Teaching and transfer
1. Each interior entry has two parents
The edge entries are one. Every interior entry is found by adding the two entries directly above-left and above-right.
Worked example A: Build Row 4
Starting from Row 3 = 1,3,3,1, keep one at each edge. The interior entries are 1+3=4, 3+3=6 and 3+1=4. So Row 4 is 1,4,6,4,1.
Worked example B: Build Row 6
From Row 5 = 1,5,10,10,5,1, add adjacent pairs to obtain 6,15,20,15,6. Hence Row 6 is 1,6,15,20,15,6,1.
Worked example C: Recover a missing entry
If an interior entry is 10 and one parent is 4, the other parent must be 6 because 4+6=10. The local addition rule can therefore be used forwards or backwards.
Row length
Row n has n+1 entries under our Row-0 convention. Row 6 has seven entries.
2. The triangle contains several nested patterns
The first diagonal is all ones. The second diagonal contains 1,2,3,4,5,… . The third diagonal contains the triangular numbers 1,3,6,10,15,… .
Worked example D: Row sums
Row 0 sums to 1, Row 1 to 2, Row 2 to 4, Row 3 to 8 and Row 4 to 16. Each row sum doubles. Thus Row 6 sums to 64.
This doubling can be understood through selection: from n objects, each object either is chosen or is not chosen, giving 2×2×…×2 = 2^n subsets.
Worked example E: Symmetry
Row 5 is 1,5,10,10,5,1. The symmetry reflects a counting fact: choosing two objects to include from five is equivalent to choosing the three objects to leave out. Both counts are ten.
Worked example F: Triangular-number diagonal
The entries 1,3,6,10,15 occur in the third diagonal. They match the triangular numbers because choosing two items from an increasing number of objects counts pairs, and the number of new pairs grows by 1,2,3,4,… .
Patterns are evidence to explain, not just decorations to notice
A numerical pattern becomes mathematically useful when we can connect it to the construction rule or to a counting interpretation. “The row looks symmetric” becomes stronger when paired with a one-to-one relation between selecting r objects and leaving out n−r.
3. Combinations count selections when order does not matter
If we choose two pupils from A,B,C,D, the pair AB is the same selection as BA. The unordered pairs are AB,AC,AD,BC,BD,CD: 6 pairs. This number appears in Row 4 of Pascal’s Triangle.
Worked example G: Choose one from five
There are five possible one-person selections. Row 5 contains the entry five next to each edge.
Worked example H: Choose two from five
The ten pairs are AB,AC,AD,AE,BC,BD,BE,CD,CE,DE. So choosing two from five gives 10.
Worked example I: Choose three from five
There are also ten. One way to see this is complement reasoning: choosing three to include is equivalent to choosing two to exclude.
Worked example J: Choose none or all
There is exactly one way to choose none and one way to choose all. These are the edge ones of every row.
Order changes the problem
Choosing two captains from A,B,C,D as an unordered pair gives six outcomes. Choosing a captain and vice-captain gives twelve ordered outcomes because AB and BA now have different roles. Pascal’s Triangle answers the unordered selection question.
Worked example K: Three toppings from six
If order does not matter and no topping is repeated, the number of three-topping selections from six is the central entry of Row 6: 20.
4. Shortest grid paths are combinations in disguise
Suppose a shortest route from (0,0) to (3,2) may move only one unit right R or one unit up U. Every shortest route must contain exactly three R moves and two U moves, in some order. So each path corresponds to choosing which two of the five move positions are U. There are 10 shortest paths.
Worked example L: Two right, two up
From (0,0) to (2,2), each shortest path contains two R and two U moves. Choose which two of four positions are U: 6 paths.
Worked example M: Four right, one up
Five moves total, with one U. Choose its position: 5 paths.
Worked example N: Three right, three up
Six moves total; choose the positions of three U moves. Row 6 central entry gives 20 paths.
Worked example O: A required intermediate point
Count shortest paths from (0,0) to (4,3) that must pass through (2,1). There are 3 shortest paths from (0,0) to (2,1): arrange R,R,U. From (2,1) to (4,3), there are 6: arrange R,R,U,U. Multiply independent choices: 3×6=18 paths.
Worked example P: Avoid one point by subtraction
There are 10 shortest paths from (0,0) to (3,2). How many pass through (1,1)? To reach (1,1), there are 2 paths. From (1,1) to (3,2), arrange two R and one U: 3 paths. Thus 6 pass through the point and 4 avoid it.
5. The addition rule has a counting explanation
Why does 4+6=10 appear in the triangle? Consider choosing two objects from five where one special object is named S. Every selection either includes S or does not.
If it includes S, choose the remaining one object from the other four: 4 ways. If it does not include S, choose both objects from the other four: 6 ways. The two cases are disjoint and complete, so total=4+6=10.
Worked example Q: Choose three from six
Name one special object. Selections including it correspond to choosing two from the other five: 10 ways. Selections excluding it correspond to choosing three from the other five: also 10 ways. Total 20. This explains the central 10+10=20 in Row 6.
Worked example R: Path version
At an interior grid point, every shortest path arriving there must come from exactly one of two predecessor points: from the left or from below. Therefore the number of paths to the point equals the sum of the path counts to those two predecessors. A Pascal-style grid grows naturally.
Worked example S: Boundary counts stay one
Along the bottom edge, if only right/up moves are allowed, there is exactly one shortest path to each bottom-edge point: move right every time. The same holds along the left edge with up moves. These boundary ones match the triangle’s edge ones.
Worked example T: Exact counting versus listing
For small cases, listing six paths is manageable. For six right and six up, listing becomes cumbersome, while Pascal’s Triangle gives the central Row-12 count 924. Structure replaces exhaustive enumeration.
6. Practice: 24 original questions
Use the convention that the top 1 is Row 0.
Questions 1–8: Build and read Pascal’s Triangle
1. Write Rows 0 through 4.
2. Build Row 5 from Row 4.
3. Build Row 6 from Row 5.
4. What is the sum of Row 4?
5. What is the sum of Row 6?
6. How many entries are in Row 8?
7. In Row 6, what is the middle entry?
8. Explain why every row begins and ends with 1 in the selection interpretation.
Questions 9–16: Combinations
9. How many unordered pairs can be chosen from four people?
10. How many unordered pairs can be chosen from five people?
11. How many groups of three can be chosen from five people?
12. How many groups of three can be chosen from six people?
13. How many groups of four can be chosen from six people?
14. How many ways are there to choose one object from seven?
15. How many ways are there to choose no objects from seven?
16. Why is choosing two from five counted the same as choosing three from five?
Questions 17–24: Shortest paths
17. From (0,0) to (2,2), moving only right or up, how many shortest paths are there?
18. From (0,0) to (3,2), how many shortest paths?
19. From (0,0) to (4,1), how many shortest paths?
20. From (0,0) to (3,3), how many shortest paths?
21. From (0,0) to (4,3), how many shortest paths pass through (2,1)?
22. From (0,0) to (3,2), how many shortest paths pass through (1,1)?
23. For Question 22, how many shortest paths avoid (1,1)?
24. A shortest route uses six right moves and six up moves. How many distinct shortest routes are possible?
7. Worked answers
Answers 1–8
1. Row 0: 1; Row 1: 1 1; Row 2: 1 2 1; Row 3: 1 3 3 1; Row 4: 1 4 6 4 1.
2. 1,5,10,10,5,1. Add adjacent entries in Row 4.
3. 1,6,15,20,15,6,1.
4. 16. 1+4+6+4+1.
5. 64. Row sums follow 2^n, so 2^6=64; direct addition agrees.
6. 9 entries. Row n has n+1 entries.
7. 20. Row 6 is 1,6,15,20,15,6,1.
8. There is exactly one way to choose none of n objects and exactly one way to choose all n.
Answers 9–16
9. 6. AB,AC,AD,BC,BD,CD.
10. 10. This is the third entry of Row 5.
11. 10. Symmetric with choosing two to leave out.
12. 20. Central entry of Row 6.
13. 15. Choosing four to include is equivalent to choosing two of six to exclude.
14. 7.
15. 1. The empty selection is unique.
16. Every chosen pair determines exactly the complementary group of three not chosen, and vice versa.
Answers 17–24
17. 6. Arrange two R and two U moves.
18. 10. Choose the positions of two U moves among five.
19. 5. Choose where the single U appears among five moves.
20. 20. Arrange three R and three U moves.
21. 18. Three paths to (2,1) and six from there to (4,3); 3×6.
22. 6. Two ways to reach (1,1), then three ways to reach (3,2).
23. 4. Ten total minus six through the point.
24. 924. Choose the positions of six U moves among twelve; the central entry of Row 12 is 924.
8. Teaching and transfer
If a learner copies rows without understanding, cover the next row and ask which two parent entries create one interior value. The triangle becomes a local rule rather than a memorised picture.
When order is accidentally counted
Ask whether AB and BA represent different outcomes. For an unordered pair they do not; for captain and vice-captain they do. The wording of the selection controls the count.
When path counting becomes a drawing marathon
Write the required moves first. Three right and two up means every shortest route is an arrangement of five symbols with two U positions. The spatial problem compresses into a selection problem.
When symmetry is treated as coincidence
Pair each selection of r objects with its complement of n−r objects. This one-to-one pairing explains why opposite entries in a row are equal.
When the Pascal addition rule feels magical
Split every selection into two non-overlapping cases: includes the special object or excludes it. Or split every path by its final step. The two-parent addition rule is then a count of complete cases.
Connect to systematic listing
Use listing for small cases as a verification tool, then let Pascal’s Triangle replace listing when the cases become numerous. This preserves the completeness habit from Systematic Listing.
Continue through this enrichment collection
For worst-case guarantees, use Pigeonhole Principle, Guaranteed Repetition and Distribution. For fraction decompositions, use Egyptian Fractions, Unit Fractions and Decomposition. For geometric rearrangement, use Geometric Dissections, Rearrangement and Area Proofs.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Row indexing is stated explicitly; formal combination notation is optional enrichment.