You have eight different suits to choose from to take on a trip. How many combinations of three suits could you take?

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You have eight different suits to choose from to take on a trip. How many combinations of three suits could you take?

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Answer 1 · 2017-05-27T07:59:49

$C_{8, 3} = \frac{8!}{(3!) (8 - 3)!} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 5!} = 56$ $C_(8,3)=(8!)/((3!)(8-3)!)=(8!)/(3!5!)=(8xx7xx6xx5!)/(3xx2xx5!)=56$

Explanation:

We can use the combinations general formula:

$C_{n, k} = \frac{n!}{(k!) (n - k)!}$ $C_(n,k)=(n!)/((k!)(n-k)!)$ with $n = population, k = picks$ $n="population", k="picks"$

and so

$C_{8, 3} = \frac{8!}{(3!) (8 - 3)!} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 5!} = 56$ $C_(8,3)=(8!)/((3!)(8-3)!)=(8!)/(3!5!)=(8xx7xx6xx5!)/(3xx2xx5!)=56$

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You have eight different suits to choose from to take on a trip. How many combinations of three suits could you take?

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