For any integer k > 1, k! is...?

Question

For any integer k > 1, k!, is the product of every integer from 1 to k. Thus, 3! = 1 x 2 x 3. What's the only integer n>32 that satisfies (n-23)!(23)! = (n-32)!(32)! ? Any help would be appreciated, thanks.

Answer 1

This is simple:

(n - 23)!(23)! = (n - 32)!(32)!

Write it like this:

1/[(n - 32)!(32)!] = 1/[(n - 23)!(23)!]

Now multiply both sides by n!:

n!/[(n - 32)!(32)!] = n!/[(n - 23)!(23)!]

Do you notice what those are now?  Yup, they are combinations.  Therefore, it is equivalent to:

nC₃₂ = nC₂₃

Remember that we have that:

nCk = nC(n - k)

Therefore, we get two equations:

k = 32

n - k = 23

From which we find that n = 55.

Hope this helps!