Practice Maths

Pascal’s Triangle and Combinations

Key Terms

The combination nCr (read “n choose r”) counts the number of ways to select r items from n items without regard to order.
Formula: nCr = n! / (r!(n − r)!), where n! = n × (n−1) × … × 2 × 1 and 0! = 1.
Pascal’s triangle
lists the values of nCr row by row: row n gives nC0, nC1, …, nCn.
Symmetry
nCr = nCn−r — choosing r to include is the same as choosing n−r to exclude.
Pascal’s rule
nCr = n−1Cr−1 + n−1Cr — each entry is the sum of the two entries above it.
Special values: nC0 = 1, nC1 = n, nCn = 1.

Pascal’s Triangle (rows 0 to 5)

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 n=0 n=1 n=2 n=3 n=4 n=5
PropertyFormula
DefinitionnCr = n! / (r!(n−r)!)
SymmetrynCr = nCn−r
Pascal’s rulen−1Cr−1 + n−1Cr = nCr
Row sumnC0 + nC1 + … + nCn = 2n
Hot Tip Use the symmetry property to simplify: 10C8 = 10C2 = 45 is much faster than computing 8! in the denominator directly.

Worked Example 1 — Evaluating nCr

Question: Evaluate 6C2.

Method: 6C2 = 6! / (2! × 4!) = (6 × 5) / (2 × 1) = 30/2 = 15.

Check (Pascal’s triangle): Row 6, position 2: 15. ✓

Worked Example 2 — Applying Pascal’s Rule

Question: Use Pascal’s rule to find 7C3 from row 6 values.

Solution: 7C3 = 6C2 + 6C3 = 15 + 20 = 35.

Pascal’s Triangle: A Pattern of Patterns

Pascal’s triangle is generated by starting with 1 at the apex, then placing 1s along both edges. Each interior entry is the sum of the two entries directly above it. Rows are numbered from row 0 (just “1”) to row n, and each row n contains the binomial coefficients nC0, nC1, …, nCn.

Beyond the obvious pattern, Pascal’s triangle contains a remarkable number of hidden structures. The row sums double each time: 1, 2, 4, 8, 16, 32 — each row n sums to 2n. The diagonal of 1s along each edge, the diagonal of natural numbers (1, 2, 3, 4, …), and the diagonal of triangular numbers (1, 3, 6, 10, …) are all present. The Fibonacci sequence also appears in a diagonal pattern. These connections reveal why Pascal’s triangle is so fundamental to combinatorics and algebra.

Combinations: Counting Unordered Selections

The combination formula nCr = n! / (r!(n − r)!) counts the number of ways to choose r items from n distinct items when order does not matter. The word “unordered” is the key distinction.

Why do we divide by r!? Because when we arrange r items in order, we have r! different arrangements of the same selection. For example, choosing {A, B, C} from a group is the same combination whether we select A then B then C, or C then B then A. There are 3! = 6 different orderings of {A, B, C}, so we divide the total number of ordered arrangements (permutations) by 6 to count only unordered selections. This is why nCr = nPr / r! = n! / (r!(n−r)!).

Pascal’s Rule: The Algebraic Proof

Pascal’s rule states: n-1Cr-1 + n-1Cr = nCr. This explains why each entry in the triangle is the sum of the two above it. The algebraic proof is illuminating:

n-1Cr-1 + n-1Cr = (n−1)!/((r−1)!(n−r)!) + (n−1)!/(r!(n−r−1)!)

Take the common denominator r!(n−r)! and simplify the numerator: (n−1)!×r + (n−1)!×(n−r) = (n−1)!(r + n − r) = (n−1)!×n = n!. So the sum equals n!/(r!(n−r)!) = nCr. ✓

There is also a beautiful counting argument: imagine a group of n people and you want to choose r. Fix one person, call them Alex. Either Alex is in the group (choose r−1 from the remaining n−1, giving n-1Cr-1 ways) or Alex is not (choose r from the remaining n−1, giving n-1Cr ways). Adding both cases gives all possible selections.

Symmetry Property: Choose to Include or to Exclude

The symmetry property nCr = nCn-r has an elegant combinatorial explanation. Choosing r items to include from n is equivalent to choosing which n−r items to leave out. Each selection of r items to include automatically determines which n−r items are excluded, establishing a perfect one-to-one correspondence between the two types of selections.

Algebraically, substitute (n−r) for r in the formula: nCn-r = n!/((n−r)!(n−(n−r))!) = n!/((n−r)!r!) = nCr. The symmetry is exact.

Practically: 10C8 = 10C2 = 45. Computing 10C8 directly requires 8! in the denominator; using the symmetry property to compute 10C2 = (10×9)/(2×1) = 45 is far faster.

Connection to the Binomial Theorem

Why do binomial coefficients (the entries in Pascal’s triangle) appear when expanding (x + y)n? Consider (x + y)² = (x + y)(x + y). When we expand, we choose x or y from the first factor and x or y from the second. The coefficient of x1y1 is 2 because there are 2C1 = 2 ways to pick one y from two factors. This counting argument scales up to any power n: the coefficient of xn-ryr in (x + y)n is nCr, because that counts the number of ways to choose r factors from which to take y (and take x from the remaining n−r factors).

Exam Tip: nC0 = 1 and nCn = 1 for all n, because there is exactly one way to choose nothing and one way to choose everything. Similarly, nC1 = n. These special cases are worth memorising to avoid unnecessary factorial calculations in exam time pressure.
Exam Tip: Do not confuse nCr (combinations, order does not matter) with nPr (permutations, order matters). For choosing a committee of 3 from 10 people, order does not matter — use 10C3 = 120. For arranging 3 runners in 1st/2nd/3rd from 10, order matters — use 10P3 = 720. The difference is a factor of r! = 6.

Mastery Practice

See Answers ➔

  1. Evaluate each combination. Show working. Fluency

    1. (a) 5C2
    2. (b) 7C0
    3. (c) 6C6
    4. (d) 8C3
    5. (e) 10C4
    6. (f) 9C7 (use symmetry)

  2. Pascal’s triangle. Fluency

    1. (a) Write out row 6 of Pascal’s triangle.
    2. (b) Use Pascal’s rule (not the formula) to write out row 7.
    3. (c) Find the sum of all entries in row 5 and in row 6. State the pattern.

  3. Use the symmetry property nCr = nCn−r to simplify, then evaluate. Fluency

    1. (a) 10C8
    2. (b) 12C10
    3. (c) 15C13
    4. (d) 20C18

  4. Simplify each factorial expression without a calculator. Fluency

    1. (a) 5!/3!
    2. (b) 8!/(6! × 2!)
    3. (c) (n+1)!/n!
    4. (d) n!/(n−2)! (express in terms of n)

  5. Use Pascal’s rule to find each value, given that 8C3 = 56 and 8C4 = 70. Fluency

    1. (a) 9C4
    2. (b) 9C5
    3. (c) 10C5 (apply Pascal’s rule twice, using your answers above)

  6. Solve for the unknown. Understanding

    Algebraic combinations.
    1. (a) nC2 = 15. Find n.
    2. (b) nC1 = 7. Find n.
    3. (c) 6Cr = 15. Find the two possible values of r.
    4. (d) nC3 = 10. Find n.

  7. Counting selections. Understanding

    Real-world combinations.
    1. (a) A team of 4 is chosen from 9 people. How many different teams are possible?
    2. (b) A committee of 3 is chosen from 5 men and 4 women. How many committees contain exactly 2 women?
    3. (c) From a standard deck of 52 cards, how many 5-card hands contain exactly 3 aces?

  8. Properties of Pascal’s triangle. Understanding

    Discovering and applying identities.
    1. (a) Show that nC0 + nC1 + nC2 + … + nCn = 2n by verifying this for n = 3, 4, and 5.
    2. (b) The alternating sum nC0 − nC1 + nC2 − … ± nCn. Evaluate this for n = 3, 4, 5. What is the pattern?

  9. Proving combination identities. Problem Solving

    Algebraic proof using the formula. Show each identity by expanding with the factorial formula.
    1. (a) Prove that nCr = nCn−r algebraically.
    2. (b) Prove Pascal’s rule: n−1Cr−1 + n−1Cr = nCr. (Hint: write both fractions with a common denominator r!(n−r)! and simplify the numerator.)

  10. Advanced counting with combinations. Problem Solving

    Multi-step counting problems.
    1. (a) A class of 12 students needs to be divided into two groups of 6. How many ways can this be done? (Hint: if the groups are unlabelled, divide by 2 to avoid double-counting.)
    2. (b) How many ways can you choose a president, vice-president, and a 3-person committee from a club of 10 people, if the president and vice-president cannot be on the committee? (Hint: do the officer selections first.)
    3. (c) The number of diagonals of a convex polygon with n sides is given by D = nC2 − n. Explain why, then find the number of diagonals in a convex decagon (10 sides).