# Binomial Coefficient - Factorials Part II

1. Jul 24, 2013

### reenmachine

1. The problem statement, all variables and given/known data

This one is trickier than the problem in my other thread in my opinion.Twenty-one people are to be divided into two teams , The Red Team and the Blue Team.There will be 10 people on Red Team and 11 people on Blue Team.How many ways to do this?

I am not sure how to solve this problem but I have the feeling it is simply $\binom{21}{10}$ or $\binom{21}{11}$ which are the same thing since: $\frac{21!}{10!(21-10)!} = \frac{21!}{10!11!}$ and $\frac{21!}{11!(21-11)!} = \frac{21!}{11!10!}$.

Now since 21! is a huge number , I decided to use a big number calculator online.Hopefully this is not forbidden.

2. Relevant equations

$\frac{21!}{11!(21-11)!} = \frac{51 090 942 171 709 440 000}{39916800 \cdot 3628800} = \frac{51 090 942 171 709 440 000}{144 850 083 840 000} = 352 716$

thoughts on this one?

thank you!!!!

Last edited: Jul 24, 2013
2. Jul 24, 2013

### haruspex

Right again.
Spreadsheet software has functions for this sort of thing. In OpenOffice Calc you'd write =combin(21;10).

3. Jul 24, 2013

### reenmachine

Thank you!

Unfortunately I am not familiar with spreadsheet software or OpenOffice Calc.

4. Jul 25, 2013

### Seydlitz

I think formally it can be done like this: $\binom{21}{10}\times\binom{11}{11}$

5. Jul 25, 2013

### Ray Vickson

If you use software that does exact rational calculations, you can do it more easily as follows:
$${n \choose k} = \frac{n(n-1)(n-2) \cdots (n-k+1)}{k (k-1) (k-2) \cdots 1} = \frac{n}{k} \cdot \frac{n-1}{k-1} \cdot \frac{n-2}{k-2} \cdots \frac{n-k+1}{1} \\ = \prod_{j=0}^{k-1} \frac{n-j}{k-j}$$
$${21 \choose 10} = \frac{21}{10} \cdot \frac{20}{9} \cdot \frac{19}{8} \cdots \frac{12}{1}$$