# Combination Problem

• DecayProduct

C(n,3) = C(n,8)

## Homework Equations

n!/(n-3)!3! = n!/(n-8)!8!

## The Attempt at a Solution

My attempts at a solution are many and varied. But, I figure my problem is with the algebraic operations on factorials. Logically, since the numerators are both n!, I figure that I can ignore that and just work on (n-3)!3! = (n-8)!8!

I boiled this down to (n-3)!/(n-8)! = 6720. I solved this based on the fact that I know that 8!/3! = 6720. Therefore, (n-3)! = 8!, and n = 11. But this was simply guess work on my part. What I'm confused about is, what is the proper way to solve the problem? How do I perform algebra on these factorials?

well, if u want a full on algebraic way then u basically write (n-3)! as

(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)!

now cancel the (n-8)! from the numerator and the denominator...

u get a fifth degree equation and u solve that!

One way would be to write the factorials in their expanded forms.
Like (n-3)! = (n-3)(n-4)(n-5)(n-6)(n-7)(n-8)!

C(n,3) = C(n,8)

Hi DecayProduct! I'd just write the answer without giving a reason …

C(n,a) = C(n,b) only if a + b = n seems too obvious to require explanation.

(If you want an explanation, I suppose you could easily show that C(n,a) is monotone increasing in a until half-way, and then monotone decreasing)