<s>[INST] 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??
 [/INST] Write it as:
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}=\frac{(x^4+x^3+x+1)^{15}}{x^{30}}$$
So, it is equivalent to:
$$(x^4+x^3+x+1)^{15}=(x^3+1)^{15}(x+1)^{15}=\\
(x^{45}+15x^{42}+\cdots+15x^3+1)(x^{15}+15x^{14}+\cdots+15x+1)=\\
x^{60}+30x^{59}+\cdots15x+1.$$
Hence, there are $61$ distinct powers of $x$.
Note: You can not use the formula you mentioned directly, because the given polynomial is lacking one term:
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}=\left(x^{-2}+x^{-1}+\color{red}{\require{cancel} \cancel{x^0}}+x+x^2\right)^{15}.$$
</s>