2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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: $$ \frac{(n+2)!}{(n-2)!4!}+\frac{(n+2)!}{(n-3)!5!}=\binom{n+2}{4}+\binom{n+2}{5}=\binom{n+3}{5}=\frac{(n+3)!}{(n-2)!5!}. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.