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Binomial Coefficient - Factorials Part II

  1. Jul 24, 2013 #1

    reenmachine

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    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. jcsd
  3. Jul 24, 2013 #2

    haruspex

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    Right again.
    Spreadsheet software has functions for this sort of thing. In OpenOffice Calc you'd write =combin(21;10).
     
  4. Jul 24, 2013 #3

    reenmachine

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    Thank you!

    Unfortunately I am not familiar with spreadsheet software or OpenOffice Calc.
     
  5. Jul 25, 2013 #4
    I think formally it can be done like this: ##\binom{21}{10}\times\binom{11}{11}##
     
  6. Jul 25, 2013 #5

    Ray Vickson

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    If you use software that does exact rational calculations, you can do it more easily as follows:
    [tex] {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}[/tex]
    So, in your case
    [tex] {21 \choose 10} = \frac{21}{10} \cdot \frac{20}{9} \cdot \frac{19}{8} \cdots \frac{12}{1} [/tex]
     
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