Question

11 people enter a pie eating contest. how many ways can the prize be awarded to the top 3

Answer 1

This is a permutation question because we care about the order.
We can demonstrate this by letting each person be a person in the pie eating contest.

A   B   C   D   E   F   G   H   I   J   K

_   _   _

Now, there are 11 ways for the first prize to be won, since there are no restrictions upheld. Let's say A wins the first prize.

B   C   D   E   F   G   H   I   J   K

A   _   _

Now, assuming prizes aren't shared, there are only ten people left to win the second prize.
Using this logic, then we can say that nine can win the third prize.

Thus, our answer is 11 · 10 · 9 = 990 ways.

However, this method works for this question.
What happens when the number of places we want gets significantly larger?
That's when we introduce the permutation formula.

We know that 11·10·9·8·7·6·5·4·3·2·1 = 11!, but we don't want 8! of them.
This is the formula for permutation.

$^{11}P_3 = \frac{11!}{(11 - 3)!} = 990$