Question

number of terms in the monomial $(xy+yz+xz)^n$

Answer 1

$\frac{(n+1)(n+2)}{2}$

Answer 2

The expression is $(xy+yz+xz)^n$. A general term in its expansion is of the form $(xy)^p (yz)^q (xz)^r$ where $p+q+r=n$. This simplifies to $x^{p+r} y^{p+q} z^{q+r}$. Each unique triplet of exponents $(p+r, p+q, q+r)$ corresponds to a distinct term. It can be shown that the mapping from $(p,q,r)$ to $(p+r, p+q, q+r)$ is one-to-one. Therefore, the number of distinct terms is equal to the number of non-negative integer solutions to $p+q+r=n$. This is a stars and bars problem with $n$ "stars" and $k=3$ "bins", given by the formula $\binom{n+k-1}{k-1}$. Substituting $k=3$, we get $\binom{n+3-1}{3-1} = \binom{n+2}{2} = \frac{(n+1)(n+2)}{2}$.