# Binomial Coefficients

I want to get ahead in my classes and learn Binomial Theorem ahead of time. What I know so far is that the formula below is the Binomial Coefficient:

$\binom n k = \frac {n!} {(n-k)!k!}$

and that you can sub in n and k for any number. However I am stuck on this particular problem:

Simplify the following into one binomial coefficient:

$\frac {(n+2)!} {(n-2)!4!} + \frac {(n+2)!}{5!(n-3)!}$

Unfortunately the textbook gives no explanation on how to tackle this problem, so I dont even know where and how to begin. Could someone point me in the right direction? Thanks in advance

• Ever heard of Pascal's identity? – Daniel W. Farlow Mar 13 '15 at 6:28
• @crash Unfortunately I haven't – user3170251 Mar 13 '15 at 6:29
• Do you mean to have $(n-3)!$ perhaps? – Daniel W. Farlow Mar 13 '15 at 6:31
• @crash yes, ill will fix it – user3170251 Mar 13 '15 at 6:32

This seems like a particularly bad problem to try to do algebraically because $k=5$, but $n$ is not determined. You said you have not heard of Pascal's identity. Here it is: $$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}\qquad\text{where}\qquad 1\leq k\leq n+1.\tag{1}$$ Looking at your problem algebraically, it appears to actually be easier to prove $(1)$ [even an algebraic proof of $(1)$ is pretty straightforward] and then use it instead of trying to fiddle around with a bunch of factorial manipulations.
Thus, we use $(1)$ to solve your problem: $$\frac{(n+2)!}{(n-2)!4!}+\frac{(n+2)!}{(n-3)!5!}=\binom{n+2}{4}+\binom{n+2}{5}=\binom{n+3}{5}=\frac{(n+3)!}{(n-2)!5!}.$$
• @user3170251 I got the answer by using the identity given in $(1)$ [Pascal's Identity]. Try it for yourself. – Daniel W. Farlow Mar 13 '15 at 20:47