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 there are 8 basketball players on a team. only 5 can play on the court at 1 time. how many different combinations of players can the coach make. the book said the ans. was 56 I do not know how that was please show me how to solve it. thanks Lisa

Hi Lisa,

This is the number of ways of choosing 5 players from 8 players, or as it is sometimes called, 8 choose 5. Your textbook may designate n choose r by nCr or and have a formula

nCr = n!/[r! (n-r)!]

but I don't like formulas so how do you get 8C5 = 56?

First notice that 8 choose 5 is the same as 8 choose 3 since choosing five players to be on the court is equivalent to choosing 3 players to sit on the bench. I'm going to work with 8 choose 3 since it is easier.

Suppose that all 8 players are on the court and the coach is going to select 3 to sit on the bench. She can select the first player to sit on the bench in 8 ways. Now that one player is on the bench she can select the second player to sit on the bench in 7 ways and once this is done she can select the third player to sit on the bench in 6 ways. Thus she can select the three players to sit on the bench in

8 7 6 ways.

But this is too large because it took order into account. Choosing Mary and then Jo and then Sarah leave the same five players on the court as choosing Mary and then Sarah and then Jo and this is the same as choosing Jo and then Mary and the Sarah. In fact she could have selected Mary, Sarah and Jo in 6 different ways. 3 ways to select the first of them and then having selected the first, 2 ways to select the second followed by 1 way to select the third. Hence there are 3 2 1 = 6 ways to select Mary, Sarah and Jo. Thus the number of ways of selecting three players to sit on the bench is

(8 7 6)/(3 2 1) = 56

If you now look back at the formula you can see that this is what it gives also

8C3
= 8!/[3! (8-3)!)
= (8 7 6 5 4 3 2 1)/[(3 2 1)(5 4 3 2 1)]
= (8 7 6)/(3 2 1) = 56

I hope this helps,
Penny

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