This is the solution provided:
$$C(k+n-1,n) \times n!$$
I know that $C(k+n-1,n)$ denotes the number of ways n indistinguishable books can be placed on k distinguishable shelves. But I'm not too sure how multiplying that by $n!$ will give us the solution to the question :/
On a side note, when we are given a question that requires us to arrange n distinguishable objects into k distinguishable bins, is it correct to say that to obtain the answer we simply just have to multiply n! to the number of ways required to arrange n indistinguishable objects into k distinguishable bins?
Best Answer
There are $n!$ ways to choose an order for the books. Once chosen this order, we have to place $k-1$ bars between the books to determine where ends a shelf and begins the next one. This is like rearrange a set of $n$ books and $k-1$ bars, and there are $\binom{n+k-1}{n}$ ways to do this.