# Solving equations in binomial coefficients like $\binom {16}{x+1}+\binom {16}{x+2}=50$

I'm trying to solve equations that have binomial coefficients (or combinations) inside. Like this one:

$$\binom {16}{x+1}+\binom {16}{x+2}=50$$

I need to LEARN how to do them. I googled but I can't seem to find anything. Is there any easier way than just making the whole thing with the formula?

Thanks a lot

Recall that $\binom n r$ is the number of ways of choosing $r$ objects out of $n$ objects and it can be shown that, $$^nC_r :=\binom n r=\dfrac {n!}{r!(n-r)!}$$

Firstly, prove that $$\binom n r + \binom n {r+1}=\binom {n+1} {r+1}$$

Now, what happens to your equation?

You have that \begin{align}\binom {17}{x+3}&=50\\\dfrac{17!}{(x+3)!(14-x)!}&=50\end{align}

Now we need to use number theoretic arguments to get the answer:

Note that as there is a factor of $25$ in RHS, there must be a factor of $25$ in LHS, which means the following:

$x+3<10 \implies x<7$ and $14-x<10 \implies x>4$. But, then that would mean that, $4<x<7$ and ass $\binom n r = \binom n {n-r}$ we'll have that $\binom {17} 9 = \binom {17} 8$, but none of which equal $50$ and hence no integral solutions.

However, there are solutions in real numbers using the generalized factorial (Gamma, $\Gamma$) functions!

• I think that that is the way to go, thanks a ton!! – Gaspa79 Feb 8 '12 at 17:35
• BTW, the answer is $x=5,6$. Did you get that? Or I'll add some extra details. Anyway, I'll do it because, people might argue, there are some number theoretic arguments here and therefore, this os not of much help! – user21436 Feb 8 '12 at 17:43
• @KannappanSampath Can I contact you in the chat? – Pedro Tamaroff Feb 8 '12 at 18:04
• @KannappanSampath Just have to ask you something. – Pedro Tamaroff Feb 8 '12 at 18:08
• Good @Gaspa79. You'd learn better! – user21436 Feb 8 '12 at 18:33