1. ## Newton's binomial

I need to prepare for a math exam, however I haven't looked at anything resembling an equation in 10 years. I would appreciate any help you can provide, as I have lots of catch-up to do.

$\displaystyle ^{n+3} C _{n}~-~^{n+2} C _{n-1}~=~15(n+1)$

How can I find n?

Probably I should use Newton's binomial, but I can't figure out how.

2. Originally Posted by avorenus
I need to prepare for a math exam, however I haven't looked at anything resembling an equation in 10 years. I would appreciate any help you can provide, as I have lots of catch-up to do.

$\displaystyle ^{n+3} C _{n}~-~^{n+2} C _{n-1}~=~15(n+1)$

How can I find n?

Probably I should use Newton's binomial, but I can't figure out how.
The first thing you need to know is that $\displaystyle ^nC_r = {}^nC_{n-r}$. One way of seeing this is to notice that choosing r objects out of n amounts to the same thing as (not) choosing the remaining n–r objects. Algebraically, both numbers are given by the same formula $\displaystyle \frac{n!}{r!(n-r)!}$.

So $\displaystyle ^{n+3} C _{n} - {}^{n+2} C _{n-1} = {}^{n+3} C _{3} - {}^{n+2} C _{3}$, and your equation becomes

$\displaystyle \frac{(n+3)(n+2)(n+1)}6 - \frac{(n+2)(n+1)n}6 = 15(n+1)$.

Cancel the factor n+1 that occurs in each term, multiply out what's left, and you'll get a simple equation for n.