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A Football/Soccer team consists of $9$ players. $2$ wingers, $2$ midfielders, $2$ strikers and three defenders. (yes I know there's no Goalkeeper.) The coach has $11$ wingers available to him, $7$ midfielders, $5$ strikers and $9$ defenders. How many ways can the team be selected? I know to pick the wingers there are $$ \frac{11!}{2!(11-2)!} \text{ways} $$ And the same goes for the rest of the positions But how do i find the total number of ways to pick the entire team? Do I just sum them all? Thanks in advance

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    (Small typo, it should be $10$ players in the team) – 2017-01-24
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    There are nine. Nine players on the team. Count them, 2, 2, 2, and 3. Nine! Whahahahha! (Crackadoom!) – 2017-01-24
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    hahahaha sorry my bad – 2017-01-24
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    Woops. It appears I can't count. – 2017-01-24

2 Answers 2

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You want to combine all possibilities into one big team arrangement. All selections (winger, midflier...) are independant one from another, so you should multiply them : the answer to your problem is ${11 \choose 2}\times{7 \choose 2}\times{5 \choose 2}\times{9 \choose 3}$

If you want to be convinced of it, just look at this small example.
I have 2 choices for my main plate, and 3 choices for my desert. How many different menus are there?

If I pick the first main plate, I have 3 choices for desert.
If I pick the second main plate, I have 3 choices for desert.

For every choice of main plate (and there are 2 of them), I have 3 choices for desert: I actually have $2 \times 3$ choices in total. Hence, multiplication is what you're looking for.

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    What if the coach wanted a specific winger to be on the team? – 2017-01-24
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    Then, what are your remaining possibilities? – 2017-01-24
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    Remaining would be another winger, and then the same selection process. Would it be $$10\cdot {7 \choose 2}\cdot {5 \choose 2}\cdot {9 \choose 3}$$ – 2017-01-24
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    Exactly! Also, you can note that $10= {10\choose 1} $ so if you want to stick to combinatorics symbols, you can (is not necessary though, up to you) – 2017-01-24
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$${11 \choose 2}\cdot {7 \choose 2}\cdot {5 \choose 2}\cdot {9 \choose 3}$$

You have to multiply them because, for example, for every choice of the wingers you have:

$${7 \choose 2}\cdot {5 \choose 2}\cdot {9 \choose 3}$$

possibilities for the rest of the team.

And you can apply the same idea for the others positions.