Binomial Coefficient Math Example 4

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

Example 4

easy
Calculate: (a) (40)\binom{4}{0}, (b) (44)\binom{4}{4}, (c) (41)\binom{4}{1}.

Solution

  1. 1
    (a) (40)=4!0!⋅4!=11=1\binom{4}{0} = \frac{4!}{0! \cdot 4!} = \frac{1}{1} = 1 (one way to choose nothing)
  2. 2
    (b) (44)=4!4!⋅0!=1\binom{4}{4} = \frac{4!}{4! \cdot 0!} = 1 (one way to choose everything)
  3. 3
    (c) (41)=4!1!⋅3!=4×3!1×3!=4\binom{4}{1} = \frac{4!}{1! \cdot 3!} = \frac{4 \times 3!}{1 \times 3!} = 4

Answer

(40)=1\binom{4}{0}=1, (44)=1\binom{4}{4}=1, (41)=4\binom{4}{1}=4.
Edge cases: there is always exactly 1 way to choose 0 items (the empty set) and 1 way to choose all items. Choosing 1 item from n gives n ways. These boundary cases confirm the formula is consistent with intuition.

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)!}.

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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 5 hard

A committee of 3 is chosen from 8 people. How many possible committees exist? If one specific pair (

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