How many terms are in $\sum \alpha_1^{a_1}\alpha_2^{a_2}\cdots \alpha_r^{a_r}\alpha_{r+1}\alpha_{r+2}\cdots \alpha_s$

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Suppose that $\alpha_1, \cdots, \alpha_n$ be $n$ roots of the polynomial equation $p(x)=0$ of degree $n$. I was studying on symmetric polynomial and have come accross of several problems on like $\sum \alpha_i^2, \sum \alpha_i^3\alpha_j^2$ etc in terms of the coefficients of $p(x)$. Then the following question came to mind.

Suppose we consider $\sum \alpha_1^{a_1}\alpha_2^{a_2}\cdots \alpha_r^{a_r}\alpha_{r+1}\alpha_{r+2}\cdots \alpha_s$ where $a_1, \cdots, a_r>1$ and $s\leq n$. In other words, we are considering the sum of all the terms of the above form where the first $r$ $\alpha$s are of higher power and the rest are of unit power.

My question is: how to find out how many terms will be there in this sum? Is it possible to get the answer in closed form?

Thanks in advance

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The sum you are considering is called the monomial symmetric function $m(a_1,a_2,\ldots,a_r,1,\ldots,1)$. One term is determined by associating a specific exponent (possibly $0$) to each root $\alpha_i$, in such a way that each exponent is used exactly the right number of times. In other words you can consider the available exponents (again including occurrences of $0$) as $n$ letters, to be used in a word formed by rearranging them (as in Scrabble), called the multiset permutations of the multiset of letters.

The number of (distinct) multiset permutations of a given finite multiset depends only on the multiplicities $m_1,\ldots,m_k$ of the various letters that occur (not on what those letters are). It is a multinomial coefficient $$ \binom n{m_1,\ldots,m_k} \quad\text{where $n=m_1+\cdots+m_k$.} $$