Solve for a 12C5+12Ca=13C6^{12}C_5+^{12}C_a=^{13}C_6

12C5+12Ca=13C6^{12}C_5+^{12}C_a=^{13}C_6

So far I did:

12C5+12Ca=13C6=^{12}C_5+^{12}C_a=^{13}C_6 =
792+12!(12−a)!a!=1716792 + \frac{12!}{(12-a)!a!} = 1716
12!(12−a)!a!=924\frac{12!}{(12-a)!a!} = 924

What do I do next?

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1

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

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3 Answers
3

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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:
12Ca=12!(12−a)!a!^{12}C_a=\frac{12!}{(12-a)!a!} from which we will recognize what value aa 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.

13!7!6!=13C6=12C5+12Ca=12!7!5!+12!(12−a)!a!\frac{13!}{7!6!}= {}^{13}C_6={}^{12}C_5+{}^{12}C_a=\frac{12!}{7!5!}+\frac{12!}{(12-a)!a!}
So,
12!(12−a)!a!=13!7!6!−12!7!5!=13!7!6!−12!7!5!66=13⋅12!7!6!−6⋅12!7!6!=13⋅12!−6⋅12!7!6!=7⋅12!7!6!=12!6!6!\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 12!(12−a)!a!=12!6!6!\frac{12!}{(12-a)!a!}=\frac{12!}{6!6!}, and hence a=6a=6.

Why do you multiply −12!7!5!-\frac{12!}{7!5!} by 66\frac{6}{6}?
– SilenceOnTheWire
Oct 20 at 22:07

@SilenceOnTheWire To get a common denominator so that I can add the fractions. I could have cancelled a ‘6’ from 13!7!\frac{13!}{7!}, but then I lose a recognizable factorial in the numerator.
– Mathily
2 days ago

Note that 924924 is divisible by 77, so that 77 must not appear in the denominator. This leads to a unique possible solution.

Alternative solution:

From Pascal Triangle, it is observable that

(nr)+(nr+1)=(n+1r+1){n \choose r}+{n \choose r+1}={n+1 \choose r+1}

In this context, n=12n=12 and r=5r=5, so a=6a=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}\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}