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

[yourmom98]

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?

 

[0rthodontist]

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.

 

[HallsofIvy]

$$C(28, 2n)= \frac{28!}{(2n)!(28- 2n)!}$$
$$C(24,2n-4)= \frac{24!}{(2n-4)!(24- (2n-4))!}$$
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 $\frac{28!}{24!}$? What is $\frac{(2n-4)!}{(2n)!}$?
Set equal to $\frac{225}{11}$ and solve for n.