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Is there a neat analytic way to solving a problem like this?

If $\dbinom{n+1}{r-2}=28$, find the value of $(n-7)!$

Assume $n,r\in\mathbb{N}$ with $0\le r\le n+1$.

An easy way to do it would be to memorize or construct Pascal's triangle to find out that the first instance of $28$ in the array for $\dbinom jk$ occurs when $j=8$, and so for this problem $n+1=8\implies n=7$ (and $r-2=2$ or $r-2=6$, so that $r=4$). Then finding $(n-7)!$ is trivial. But I don't particularly like relying on rote memorization or table construction.

I was wondering if this would be doable with the recursive formula $\dbinom j{k-1}+\dbinom jk=\dbinom{j+1}k$. It seems to me that the argument $r-2$ complicates any attempt to decompose $\dbinom{n+1}{r-2}$ into simpler coefficients but I wouldn't know where to begin. I suspect a clever manipulation of the factorial formula might also be useful, though nothing immediately comes to mind.

user170231
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  • One answer is $n=27$ and $r=3$. Are you sure you reproduced the problem completely? – quid Sep 09 '16 at 18:54
  • @quid I'm pretty sure - I don't know what the source of the problem is, as I found it posted somewhere on a tutoring forum some time ago. I had jotted it down as I saw it with the intention of giving it a try in the future. It's entirely possible the original question was missing some extra details (maybe find the smallest $n$?). – user170231 Sep 09 '16 at 18:59

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As $\binom{m}{k}$ is a product and quotient of numbers $\le m$, and the prime $7$ divides $28$ we get that $7 \le n+1$.

Now, the fact $\binom{m}{1}= m$ always yields a trvial solution, also $\binom{m}{m-1}=m$ but this is excluded.

Ignoring the extremal cases $m=0,1,m-1,m$ we have $\binom{m}{k}\ge m(m-1)/2$.

Then $m(m-1)/2 \le 28$ yields $m \le 8$. Thus $7 \le n+1 \le 8$. And one is pretty close to having found your solution.

quid
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