# Number of terms in a monomial symmetric polynomial

Is there a closed form expression for the number of terms in a monomial symmetric polynomial in a given number of variables for a particular partition of exponents, in terms of which/how many exponents are distinct?

I feel the answer should be straightforward, and it's probably just a contrived statement of a more elementary number theoretic question, but I'm just drawing a blank at the moment. I at first thought that the answer might be

$$\frac{N!}{(N-k+1)!}$$

for $N$ variables and $k$ distinct exponents, which works for all $N=2$ and $N=3$ cases, but predicts 4 terms for a partition $\alpha=(1,1,0,0)$, while there are in fact 6:

$$m_\alpha(a,b,c,d) = ab+ac+ad+bc+bd+cd$$

Am I being an idiot?

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If you have $k_i$ exponents of value $i$, then it's the multinomial ceofficient: ${n \choose k_1,k_2,...,k_n}$. Hope I didn't mess it up. – user3533 Dec 15 '10 at 23:35
@user3533: I believe that's correct, although I'm still searching for a reference. You should post that as an answer. – Mike Spivey Dec 15 '10 at 23:40
Or more accurately: If the smallest exponent appears $k_1$ times, the second smallest appears $k_2$ times, etc. then the answers is the multinomial ceofficient: ${n \choose k_1,k_2,...,k_n}$. Hope I didn't mess it up. – user3533 Dec 15 '10 at 23:41

If the smallest exponent appears $k_1$ times, the second smallest appears $k_2$ times, etc. then the answer is the multinomial coefficient: ${n \choose k_1,k_2,...,k_n}$.

The multinomial coefficient $n \choose k_1,k_2,...,k_n$ counts how many ways you have to partition n objects into disjoint subsets of sizes $k_1,...,k_n$ which is just what you need.

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Suppose the number of indeterminates is $n$. Suppose the monomial symmetric polynomial is of type $(a_1,a_2,...,a_n)$ where the first $b_1$ exponents are identical, the next $b_2$ are identical and so on until $b_r$, where $1\leq b_j \leq n$ and $\sum_{1}^{r} b_j = n$. Then, number of terms in a monomial symmetric polynomial is the same as number of permutations of the set of exponents. The number of permutations when they are all distinct is $n!$. If not, the every permutations of any collection of identical exponents amongst themselves should be treated the same. So factoring out such permutations, we get the total number to be $\frac{n!}{b_1!\cdot b_2!\cdot...\cdot b_r!}$

For example, consider the examples on the wikipedia page. We have, $n=3$. In the first case, we have 2 identical exponents, so number of terms is $\frac{3!}{2!}=3$.

In the second case, all exponents are distinct, so the number of terms is $3!=6$

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Which is the multinomial coefficient, as given by user3533. :) – Mike Spivey Dec 15 '10 at 23:47
@Mike: Yes. When I started typing my answer, there were no comments. – Timothy Wagner Dec 15 '10 at 23:49
I have done the same thing myself many times. :) FWIW, I upvoted your answer because of the explanation. – Mike Spivey Dec 15 '10 at 23:52