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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

1 Answers 1

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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 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!

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    I think that that is the way to go, thanks a ton!!2012-02-08
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    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!2012-02-08
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    @KannappanSampath Can I contact you in the chat?2012-02-08
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    @KannappanSampath Just have to ask you something.2012-02-08
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    Actually I used ficticious numbers, the real equation was with 8 and x +1 and x+2 equals 9c3, I did this so I could learn from the question and not cheat myself. I reached the answer by the way, thanks a lot!!2012-02-08
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    Good @Gaspa79. You'd learn better!2012-02-08