Finding n and k that satisfies ${n \choose k} = 517 $ How do I find all the integers n and k that satisfies ${n \choose k} = 517 $ ? How do I approach this question?
 A: In general, finding the number of times that a number appears in Pascal's triangle is a difficult problem. There's an open problem called Singmaster's conjecture that relates to the multiplicity of entries in the triangle.

There are clearly at least two pairs: $$\binom{517}{1} \qquad \text{and} \qquad \binom{517}{516}$$

After the $517^\text{th}$ row in Pascal's triangle, every non-$1$ entry is greater than $517$, so there are a finite number of entries we can check, so it's definitely not that difficult of a problem. I'll try to write some code to check if there are any other pairs.
EDIT: @YuvalFilmus noticed that we can dramatically reduce the number of entries we need to check for non-trivial solutions to $n \leq 32$. @JorgeFernández finished the beast with $$47 \text{ prime, }\,\,\,47 > 32 \qquad \Rightarrow \qquad 47 \not\mid \binom{n}{k}$$
A: *

*${n \choose 0}$ is always 1, there is no solution

*${n \choose 1} = n$, which gives the solution ${517 \choose 1}$

*${n \choose 2} = \frac{n(n-1)}{2}$, which leads to the $n^2-n-1034=0$ quadratic equation, which doesn't have an integer solution.

*${n \choose 3} = \frac{n(n-1)(n-2)}{6}$. But ${15 \choose 3}=455$, and ${16 \choose 3}=560$, so there is no solution again.

*Similarly, hogy ${n \choose 4}$, we got ${12 \choose 4}=495$ and ${13 \choose 4}=715$.

*For ${n \choose 5}$, ${11 \choose 5}=462$ and ${12 \choose 5}=792$. Here is no solution, too.

*For ${n \choose 6}$, ${11 \choose 6}=462$ (just as above), but ${12 \choose 6}=924$.

*From this point on, if we found a solution ${n \choose k}$, we had to also find it before. (Why?)


Thus the only solutions are ${517 \choose 1}$ and ${517 \choose 516}$.
A: The idea is to use the fact that for each $n$, the binomials $\binom{n}{0},\ldots,\binom{n}{\lfloor n/2 \rfloor}$ form an increasing sequence. We can assume that $k \leq n/2$. If $k \geq 2$ then $\binom{n}{k} \geq \binom{n}{2}$, and so, since $\binom{33}{2} = 528$, we must have $n \leq 32$. Checking $\binom{n}{k}$ for all $n \leq 32$, we find no solution. Hence we are left only with the trivial solutions
$$ \binom{517}{1} = \binom{517}{516} = 517.$$
In fact, as Jorge Fernández correctly mentions, since the prime 47 is a factor of 517, we must have $n \geq 47$, and so no checking is needed.
