How to find number of distinct terms in a multinomial expansion? 
Find the number of distinct terms in the expansion of
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}$$
(with respect to powers of $x$)

I saw that the formula for the number of distinct terms (or dissimilar) in a multinomial expansion $(x_1+x_2+x_3+...+x_k)^n$ is $$\binom{n+k-1}{k-1}$$
But applying that here means $$\binom{15+4-1}{4-1}= \binom{18}{3} = 816$$
But the answer says 61.
Is there a difference between the situations for which that formula is meant to be used and that in which I am using??

Let us equivalently find the number of different monomials in 

$$
f=(x^2)^{15}\left(\frac{1}{x^2}+\frac{1}{x}+x+x^2\right)^{15}
=
\left(1 + x + x^3 + x^4\right)^{15}
$$

(after collecting them and writing the polynomial w.r.t. the basis $1,x,x^2,\dots$), which is a reciprocal polynomial of degree $4\cdot 15=60$. It is clear that there are at most $61$ terms (in the degrees $0,1,2,\dots,60$), so let us show that each degree is indeed taken. It is enough to check each degree up to $30$, the polynomial $f$ being reciprocal. We compute
$$
(x^4 + x^3 + x + 1)^2
=x^{8} + 2  x^{7} + x^{6} + 2  x^{5} + 4  x^{4} + 2  x^{3} + x^{2} + 2  x + 1
$$
and see that each degree in the range $0,1,\dots,8$ is taken. So each degree in $\color{red}1\cdot(x^4 + x^3 + x + 1)^{2\cdot 7}$, and in particular also in 
$$
f= (x^4 + x^3 + x + \color{red}1)\cdot(x^4 + x^3 + x + 1)^{2\cdot 7}$$ between $0$ and $8\cdot 7$ is taken. 
