How many distinct football teams of 11 players can be formed with 33 men? 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!
 A: 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 $$\dbinom{33}{11} = \dfrac{33 \times 32 \times 31 \times 30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24 \times 23}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 193536720$$
A: How many ways are there to choose $11$ people from $33$? Exactly $$
\binom{33}{11}.$$
A: 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.

