how many terms will there be once you collect terms with equal monomials? What is the sum of all the coefficients? If you expand $(x_1+x_2+\cdots+x_k)^n$, how many terms will there be once you collect terms with equal monomials? What is the sum of all the coefficients?
I'm kind of lost here. This came up with other questions on the topic of binomial coefficients, and i have figured those out. I have spent almost an hour trying to figure where to start and haven't made any progress.
 A: Assuming we are in a commutative ring (i.e. that $x_1x_2 = x_2x_1$ and so on),
we can use the multinomial coefficient to describe the coefficient on each distinct monomial after collecting terms.
Think of the expression $(x_1 + x_2 + \cdots+ x_k)^n$ expanded out as
$$(x_1 + x_2 + \cdots+ x_k)(x_1 + x_2 + \cdots+ x_k)\cdots(x_1 + x_2 + \cdots+ x_k) \quad\quad (n ~~times)$$
When we multiply this out, we choose one of the $x_i$ from each of the parenthetical groups $$(x_1 + x_2 + \cdots+ x_k).$$
All told, we will choose $n$ variables (some of them perhaps the same) to multiply together to create a monomial.
There are $\binom{n}{m_1,~m_2, ~m_3, \ldots, ~m_k}$ ways to get the monomial $x_1^{m_1}x_2^{m_2}\cdots x_k^{m_k}.$ This means that the coefficient is exactly that: $$\binom{n}{m_1,~m_2, ~m_3, \ldots, ~m_k} = \frac{n!}{m_1!m_2!\cdots m_n!}.$$
A: To get the sum of the coefficients, just set all the $x$'s to $1$.  What does that give you for $n=1$?  To get the number of monomials you are looking for the number of weak compositions of $n$ into $k$ parts, the number of ordered ways to write $n$ as the sum of $k$ nonnegative integers.  If you add one to each, this is the same as the number of strong compositions of $n+k$ into $k$ parts.  This is ${n+k-1 \choose k-1}$ as shown in the Wikipedia article.
