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

  • $\begingroup$ Ever heard of Pascal's identity? $\endgroup$ – Daniel W. Farlow Mar 13 '15 at 6:28
  • $\begingroup$ @crash Unfortunately I haven't $\endgroup$ – user3170251 Mar 13 '15 at 6:29
  • $\begingroup$ Do you mean to have $(n-3)!$ perhaps? $\endgroup$ – Daniel W. Farlow Mar 13 '15 at 6:31
  • $\begingroup$ @crash yes, ill will fix it $\endgroup$ – 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: $$ \frac{(n+2)!}{(n-2)!4!}+\frac{(n+2)!}{(n-3)!5!}=\binom{n+2}{4}+\binom{n+2}{5}=\binom{n+3}{5}=\frac{(n+3)!}{(n-2)!5!}. $$

  • $\begingroup$ Could you show the steps on how you got the answer (n+3)!/(n-2)!5! ? $\endgroup$ – user3170251 Mar 13 '15 at 15:11
  • $\begingroup$ @user3170251 What is the binomial coefficient of the first fraction you listed? $\endgroup$ – Daniel W. Farlow Mar 13 '15 at 19:57
  • $\begingroup$ I understand how you got (n+2, 4)+(n+2, 5), but what Im asking is how you got the answer (n+3, 5) $\endgroup$ – user3170251 Mar 13 '15 at 20:30
  • $\begingroup$ (the coeff. of the first fraction is (n+2, 4) $\endgroup$ – user3170251 Mar 13 '15 at 20:31
  • $\begingroup$ @user3170251 I got the answer by using the identity given in $(1)$ [Pascal's Identity]. Try it for yourself. $\endgroup$ – Daniel W. Farlow Mar 13 '15 at 20:47

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