Does $a!b!$ always divide $(a+b)!$ Hello the question is as stated above and is given to us in the context of group theory, specifically under the heading of isomorphism and products. I would write down what I have tried so far but I have made very little progress in trying to solve this over the last few hours! 

The number of ways to choose $a$ objects out of $a+b$ objects if order matters in the selection is $(a+b)\times(a+b-1)\times\cdots\times((a+b)-(a-1))=\frac{(a+b)!}{b!}$, since there are $a+b$ ways to choose the first object, $a+b-1$ ways to choose the second object from the remaining ones, and so on.
However, $a!$ permutations actually correspond to a single combination (where order is immaterial), since $a$ objects can be arranged in $a!$ ways. This means that $\frac{(a+b)!}{b!}=ka!$ for some integer $k$, so that $(a+b)!=ka!b!$.
