Will ${n \choose r} = {m \choose k}$ for $n \not=m$?

Question

I'm currently doing some counting techniques, and I wondered whether the follow statement is possible:

Prove or disprove ${n \choose r} = {m \choose k},$ $n \not = m$, $r \not = k$. Also assume ${n \choose r} \not= 1,$ and ${m \choose k} \not = 1$.

I thought of this statement when I did the problem ${30 \choose a} = {3a \choose b}$. The given solution was straight away, to assume $30 = 3a$.

This isn't a priority for me, so I haven't made any progress on the question. Any hints or solutions would be appreciated.

Answer 1

We have $\binom {10}3=\binom {16}2=120$.

With $\binom {30}a$ you have the large prime factor $29$ to contend with so $3a$ has to be at least $30$ (unless you count $a=b=0$ as a possible solution, $a=1$ doesn't work). Obviously $3a=30$ will give you a solution.

Answer 2

Yes, it is possible. $$\binom{14}{6} = 3003 = \binom{15}{5}$$