Find the largest values of $m$ and $n$ such that $24\ ^mC_n = \ ^{15}P_4$.

Question

The first thing I did was divide the $24$ to the other side to get $^mC_n =1365$, but I do not know what to do from there. The answer is $m=1365$ and $n=1364$.

Answer 1

Note that $4! =24$. If we divide throughout by $24$, we get $^{15}C_4$ on RHS. So, $(m,n)\equiv (15, 4),(15, 11) $ is also a solution.

Answer 2

You can demonstrate that $m=1365$ and $n=1364$ is a solution.

If $m>1365$, then either $_mC_n=0$, $_mC_n=1$, or $_mC_n>1365$. So there is no solution with $m$ greater than the $m$ in your solution. So any solution has $m\leq1365$.

Any solution must have $n<m$. So if there is a solution where $n>1364$, then it has $m>1365$. But we established there is no such solution. So any solution has $n\leq1364$.

Since $(m,n)=(1365,1364)$ is indeed a solution, it is "the largest" be several reasonable meanings for "the largest".