Question

How many ways can a person choose 3 desserts from a desserts tray 8 desserts​

Answer 1

In combinatorics, this question is a combination question, because combination deals with selection while permutations deal with arrangement.

$nCr = \frac{n!}{(n - r)!r!}$

Plugging in our values,

$8C3 = \frac{8!}{(8 - 3)!3!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6 \times 5!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2} = 56$

That means 3 desserts from a tray of 8 desserts 56 times.