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

Question

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

Answer 1

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!