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Combinations: C(28,2n)/C(24,2n-4)=225/11 Solve for n

  1. Aug 27, 2006 #1
    i am given C(28,2n)/C(24,2n-4)=225/11 where C are combonations. i am supposed to solve for n. after subbing in n and k and simplifying i get (11*28!)/2n!=(225*24!)/(2n-4)! (i am pretty sure this is correct). i do not seem to be able to continue after that. how do i solve for n?
     
  2. jcsd
  3. Aug 27, 2006 #2

    0rthodontist

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    No, a combination is
    C(n, k) = n!/(k!(n-k)!)
    it looks like you are using permutations. Once you fix that, you should look for ways to cancel out the factorials by writing them as products. n!/k! = n * (n-1) * ... * (n-k+1) with n > k will be useful.
     
  4. Aug 28, 2006 #3

    HallsofIvy

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    [tex]C(28, 2n)= \frac{28!}{(2n)!(28- 2n)!}[/tex]
    [tex]C(24,2n-4)= \frac{24!}{(2n-4)!(24- (2n-4))!}[/tex]
    so the quotient is (fortunately for us 24- (2n-4)= 24- 2n+ 4= 28- 2n)
    [tex]\frac{28!}{(2n)!(28- 2n)!}\frac{(2n-4)!(28-2n!)}{24!}

    Now, what is [itex]\frac{28!}{24!}[/itex]? What is [itex]\frac{(2n-4)!}{(2n)!}[/itex]?
    Set equal to [itex]\frac{225}{11}[/itex] and solve for n.
     
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