Binomial Coefficient Math Example 5

Follow the full solution, then compare it with the other examples linked below.

Example 5

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?

Solution

  1. 1
    Total committees: (83)=8!3!5!=8×7×63×2×1=3366=56\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56
  2. 2
    Committees with both Alice and Bob: Alice and Bob fill 2 of 3 spots; choose 1 more from remaining 6: (61)=6\binom{6}{1} = 6
  3. 3
    So 6 out of 56 committees include both Alice and Bob

Answer

Total: 56 committees. Committees with Alice AND Bob: 6.
Counting with restrictions: fix required members and count ways to fill remaining slots from the remaining pool. This technique applies to any counting problem with mandatory inclusions.

About Binomial Coefficient

The binomial coefficient (nk)\binom{n}{k} counts the number of ways to choose kk items from nn distinct items without regard to order. It equals n!k!(n−k)!\frac{n!}{k!(n-k)!}.

Learn more about Binomial Coefficient →

More Binomial Coefficient Examples

Example 1 medium

Calculate [formula] using the formula [formula], and verify by listing all combinations of 2 items f

Example 2 hard

A fair coin is flipped 5 times. Using [formula], find [formula] (exactly 3 heads).

Example 3 medium

Calculate [formula] and explain what it counts.

Example 4 easy

Calculate: (a) [formula], (b) [formula], (c) [formula].

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