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$$^{12}C_5+^{12}C_a=^{13}C_6$$

So far I did:

$$^{12}C_5+^{12}C_a=^{13}C_6 = $$ $$792 + \frac{12!}{(12-a)!a!} = 1716$$ $$\frac{12!}{(12-a)!a!} = 924$$

What do I do next?

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    $\begingroup$ Hint: Fix one of the 13 and you either choose 5 from the remaining 12, or you choose all 6 from those 12. $\endgroup$
    – NickC
    Commented Oct 20, 2016 at 18:06

3 Answers 3

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Note that $924$ is divisible by $7$, so that $7$ must not appear in the denominator. This leads to a unique possible solution.

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Alternative solution:

From Pascal Triangle, it is observable that

$${n \choose r}+{n \choose r+1}={n+1 \choose r+1}$$

In this context, $n=12$ and $r=5$, so $a=6$.

Proof of the Identity: \begin{align} {n \choose r}+{n \choose r+1}&=\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r-1)!(r+1)!}\\ &=\frac{n!(r+1)+n!(n-r)}{(n-r)!(r+1)!}\\ &=\frac{(n+1)!}{(n-r)!(r+1)!}\\ &={n+1 \choose r+1} \end{align}

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Here is a solution which is tailored to the fact that you've tagged this 'precalculus'. There are a bunch of more elegant and clever solutions that may not be appropriate to the level at which you are learning the material. The general approach will be to get something of the form: $$^{12}C_a=\frac{12!}{(12-a)!a!}$$ from which we will recognize what value $a$ takes. To do so, we will keep everything in the form of a factorial as that is the form we want to recognize the answer from.

$$\frac{13!}{7!6!}= {}^{13}C_6={}^{12}C_5+{}^{12}C_a=\frac{12!}{7!5!}+\frac{12!}{(12-a)!a!}$$ So, $$\frac{12!}{(12-a)!a!}=\frac{13!}{7!6!}-\frac{12!}{7!5!}=\frac{13!}{7!6!}-\frac{12!}{7!5!}\frac{6}{6}=\frac{13\cdot12!}{7!6!}-\frac{6\cdot12!}{7!6!}=\frac{13\cdot12!-6\cdot12!}{7!6!}=\frac{7\cdot12!}{7!6!}=\frac{12!}{6!6!}$$

Thus we see $\frac{12!}{(12-a)!a!}=\frac{12!}{6!6!}$, and hence $a=6$.

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  • $\begingroup$ Why do you multiply $-\frac{12!}{7!5!}$ by $\frac{6}{6}$? $\endgroup$ Commented Oct 20, 2016 at 22:07
  • $\begingroup$ @SilenceOnTheWire To get a common denominator so that I can add the fractions. I could have cancelled a '6' from $\frac{13!}{7!}$, but then I lose a recognizable factorial in the numerator. $\endgroup$
    – Mathily
    Commented Oct 21, 2016 at 21:10

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