Lower bound of multinomial coefficients?

Question

Multinomial coefficient $=\dfrac{n!}{a_1!\cdot a_2!\cdots a_k!}$, where $n=a_1+a_2+\cdots+a_k$.

So my thoughts are there should be a minimum when the denominator goes to the largest.

I believe there is a maximum for the denominator from my basic knowledge and experience of inequity, like for $a+b=c$, the $\max(ab)$ is reached when $a=b$, but I don't know how to do with the situation here (factorial).

Can anyone give me some help?

Thanks

Edit: Well actually, I'm using this to help me solve a problem. The answers followed here makes me feel I am trying the problem in a wrong way. Here is the question, could you tell how to think it right?

Let n larger or equal than 2. We want to select as many subsets of $[n]$ as possible, without selecting two subsets so that one of them contains the other. Prove that we can always select at least $(2^n-1)/n$ Subsets.

Thanks again

Answer 1

This is an old post but people who stumble across it a lot later might consider that all the mathematical notation and analysis of denominators and factorials can be avoided if one remembers that all multinomial coefficients--just like binomial coefficients--can be generated from a Pascal's Triangle in which the initial row is, for a $k$-nomial, a sequence of of $k$ 1's, which corresponds to $n=1$. Thus we have a recurrence relation that will generate all the multinomial coefficents for the $n$th power of a $k$-nomial. And this will, I think, suffice to show that the smallest multinomial coefficient is always 1. It will also show that the maximal coefficients are the central one(s): depending on whether $n$ is even or odd. It will show that the coefficients are symmetric around the maximal ones and so on.

For example, in the case of a trinomial $(k=2)$ we have for Pascal's Triangle,

1 1 1

1 2 3 2 1

1 3 6 7 6 3 1

1 4 10 16 19 16 10 4 1

and so on. The generalization to quadrinomials, quintinomials and higher polynomials is straightforward. Just remember how you generate the Pascal's Triangle for a binomial. Your recurrence relation for the trinomial will have additional term to the one used to generate binomial coefficients.

Again, I think this is perfectly rigorous--no further proof is needed.

NB: Mathematica has function called RecurrenceTable[] that will generate the coefficients quite quickly and efficiently so you can obtain multinomial coefficients rapidly without evaluating large factorials.

Answer 2

The lower bound of the coefficient is $1$. This happens when $a_i=n$ for some $1\le i\le k$. Then, we have $a_j=0$ for all $j \ne i$.

It must be noted that here $a_j \ge 0$ for all $1\le j\le k$. Thus, no $a_j$ in the denominator can be greater than $n$.

It can be seen that $1$ must be the minimum this way: the multinomial coefficient is the number of ways of arranging $n$ objects where $a_1$ are of one kind, $a_2$ are of a second kind, and so on. As it is the number of ways, the value must always be a positive integer for $n\ge 0$. So, as it can take the value of $1$, it must be the minimum value.

Answer 3

Note that moving from $a_i!a_j!$ with ($a_j\ge1$) to $(a_i+1)!(a_j-1)!$ increases by a factor of $\frac{a_i+1}{a_j}$, which is $>1$ whenever $a_i+1>a_j$. In particular, the denominator can be enlarged whenever there exist two nonzero arguments. Therefore, the maximal denominator can only be the one corresponding to all but one of the $a_i$ equal to zero. This gives the lower bound $1$ for the multinomial.