Question

In soccer, a tie at the end of regulation time leads to a shootout by three members from each team. How many ways can 3 players be selected from 11 players available? For 3 selected players, how many ways can they be designated as first, second, and third?

Answer 1

Answer:

C (11.3) = 165

P (3,3) = 6

Step-by-step explanation:

We want to select 3 players out of 11 regardless of the order. That is, there is no difference between selecting the players {2,5,7} or {7,2,5}

Then we use the formula of combinations:

$C(n, r) = \frac{n!}{r!(n-r)!}\\\\C(11, 3) = \frac{n!}{r!(n-r)!}\\\\C(11, 3) = 165$

There are 165 ways to choose 3 players out of 11.

Now we want to know how many ways you can designate those 3 players as first, second and third. Now if we care about the order of selection. Then we use permutations.

$P(n, r) = \frac{n!}{(n-r)!}\\\\P(3,3) = \frac{3!}{(3-3)!}\\\\P(3,3) = 6$

They can be designated in 6 different ways