Binomial Coefficient Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Binomial Coefficient.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

The number of ways to choose k items from n items, written C(n, k) or \binom{n}{k}.

Same as combination count, but now viewed as a coefficient in algebraic expansions.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The binomial coefficient bridges counting (combinations) and algebra (polynomial expansion).

Common stuck point: C(n, k) = C(n, n - k). Choosing k to include is the same as choosing n - k to exclude.

Worked Examples

Example 1

medium
Calculate \binom{6}{2} using the formula \binom{n}{k} = \frac{n!}{k!(n-k)!}, and verify by listing all combinations of 2 items from \{A, B, C, D, E, F\}.

Solution

  1. 1
    Apply formula: \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15
  2. 2
    List all 2-item combinations from \{A,B,C,D,E,F\}: AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF
  3. 3
    Count: 15 combinations ✓
  4. 4
    Confirms formula gives the correct count

Answer

\binom{6}{2} = 15. Verified by listing all 15 two-item combinations.
The binomial coefficient \binom{n}{k} counts the number of ways to choose k items from n without regard to order. Order doesn't matter in combinations (unlike permutations). The formula cancels repeated arrangements via the k! in the denominator.

Example 2

hard
A fair coin is flipped 5 times. Using P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}, find P(X=3) (exactly 3 heads).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Calculate: (a) \binom{4}{0}, (b) \binom{4}{4}, (c) \binom{4}{1}.

Example 2

hard
A committee of 3 is chosen from 8 people. How many possible committees exist? If one specific pair (Alice and Bob) must both be included, how many committees include both?
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Background Knowledge

These ideas may be useful before you work through the harder examples.

combinationfactorial