Factorial formula problem Prove that  $(n-r)!(r!)$ divides $ n! $ i know its a factorial formula and it might be easy but i stuck .I tried induction to $n$ or analyzing the factorials but im missing something
 A: You can use a combinatorial argument. How many groups of size $k$ can be formed from $k$ students? There are $n$ options for the first student, $n-1$ options for the second $\dots (n-k+1)$ options for the $k$'th student. Therefore there are $n\cdot (n-1)\cdot\dots (n-k+1)=\frac{n!}{(n-k)!}$ possible groups.
However we have overcounted! Because in a group of students there is no such thing as a first student or a second student. Every group of student has been counted exactly $k!$ times since this is the number of ways to order the $k$ students of a given group. Therefore the number of groups is $\frac{n!}{k!(n-k)!}$. Clearly the number of groups is an integer number. Then $\frac{n!}{k!(n-k)!}$ is an integer and so $k!(n-k)!$ divides $n!$
A: Here is another solution via group theory. Consider the group $S_n$, it has $n!$ elements. Now consider the subgroup of $S_n$ consisting of the permutations such that elements in $\{1,2,3\dots k\}$ always go to elements of $\{1,2,3\dots k\}$, it has order $k!(n-k)!$ so by lagrange's theorem $k!(n-k)!$ divides $n!$
