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Can anyone help me with this problem, I can't figure out how to solve it...

How many distinct football teams can be formed with 33 men?

Thanks!

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American football or proper football (or soccer, if you want to call it that)? –  Andrew D Jan 8 '13 at 2:33
    
Soccer, 11 players in one team. –  user52516 Jan 8 '13 at 2:33
3  
Are the men all distinct? –  Daenerys Naharis Jan 8 '13 at 4:20
    
You can make up to 3 distinct teams of 11 from 33 players. How many ways you can make them is an interesting question too. –  ex0du5 Jan 9 '13 at 1:59

3 Answers 3

up vote 1 down vote accepted

In general, the number of distinct ways to choose $k$ objects from $n$ objects $-$ i.e. "$n$ choose $k$" $-$ is given by the binomial coefficient: $$\binom nk = \frac{n\,!}{k\,!\,(n-k)\,!}$$

The number of distinct ways to choose $11$ players from $33$?

We simply need to compute $\,$ "$\,33$ choose $11\,$" $-$ which gives us precisely: $$\binom{33}{11} = \frac{33!}{11\,!\, (33-11)\,!}\quad\text{distinct teams that can be formed from 33 players}$$
$$\frac{33!}{11\,!\, (33-11)\,!} = \dfrac{33 \times 32 \times 31 \times \cdots \times 3 \times 2 \times 1}{(11 \times 10 \times \cdots\times 3 \times 2 \times 1)(22 \times 21 \times \cdots \times 3 \times 2 \times 1)} $$ $$\quad\quad = \dfrac{33 \times 32 \times 31 \times \cdots \times 24 \times 23}{11 \times 10 \times 9 \times \cdots \times 3 \times 2 \times 1} = 193536720$$

Hence there are 193,536,720 distinct football teams which can be formed from 33 men.


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Number of ways of choosing $r$ people from a set of $n$ people is given by $$\dbinom{n}r$$ In your case, you want to choose $11$ men from a set of $33$ men. Hence, the number of ways is $$\dbinom{33}{11} = \dfrac{33 \times 32 \times 31 \times 30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24 \times 23}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 193536720$$

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And $\binom{n}{r} = \frac{n!}{(n-r)!r!}$. That's where all the multiplications come in. (in case someone reading this doesn't know that...) –  anorton Jan 8 '13 at 2:41

How many ways are there to choose $11$ people from $33$? Exactly $$ \binom{33}{11}.$$

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