What's the explicit formula for Permutations of Subsets of a Multiset? [duplicate]

Question

What is the number of permutations of subsets of the multiset $S$ with cardinality $n$? A sample problem would be to find the number of ways you can construct "words" with three of the letters in the set $\{G,R,E,E,N\}$.

I've looked everywhere and found two identical posts by the same person on MSE and on MO, but no formula was posted. Such formula exists for sets with distinct elements, $P(n,r)$. The number of permutations of a multiset also exists as the multinomial coefficient. But there seems to be no way to compute the number of permutations of subsets of a multiset given the cardinality, other than going case by case and using logic. I've heard that generating functions can be used to solve such problems (as was written in Maple's help files) but I'm looking for a general solution using combinatorics, although a generating functions solution would be cool too (I have yet to learn generating functions).

I played around with this last week, and found a formula when the multiset only contains one element that is duplicated. Let's suppose the number of duplicates is $E$, the cardinality of the multiset $S$ is $n$, and that the cardinality of the subsets desired is $P$. Then, the number of permutations is $$\sum_{k=0}^{E}\binom{P}{k}P(n-E,P-k)$$ This formula only works when $n\geq E+P$ and even then it might not work, and I didn't know how to proceed. Suresh Venkat suggested to sum multinomial coefficients, but I don't understand his reasoning.

I originally posted this on MathOverflow, now deleted.

Answer 1

Let's define

• $S$ as the multiset,
• and $m_k$ as the multiplicity of the element $k$ of the multiset $ S=(m_1,m_2,...,m_n) $.

We want all the unique permutations of length $l$ of the multiset. For the formula, we have to find all the combinations $C=(x_1,x_2,...,x_n)$ where $0 \leq x_k \leq m_k$ such that $$ \sum_{k=1}^n x_k=l $$

What we need is actually the product of the factorials of the elements of that combination: $$ P(C)=\prod_{k=1}^n x_k! $$ Now let's define the number of combinations as J. So to answer your question, the number of permutation is $$ \sum_{j=1}^J \frac{l!}{P(Cj)} $$ or the closed-form expression, where $\binom{n}{k_1,k_2,...,k_m}=\frac{n!}{k_1!k_2!...k_m!}$ is a multinomial coefficient $$ \sum_{x_1+x_2+...+x_n=l} \binom{l}{x_1,x_2,...,x_n} $$

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Let's try this formula with this example: the multiset is $S=(3,2,1)$ and $l=4$. All the combinations are $$ C=((1,2,1),(2,1,1),(2,2,0),(3,0,1),(3,1,0)) $$ so the number of permutation is $$ \frac{4!}{1!2!1!}+\frac{4!}{2!1!1!}+\frac{4!}{2!2!0!}+\frac{4!}{3!0!1!}+\frac{4!}{3!1!0!}=\frac{24}{2}+\frac{24}{2}+\frac{24}{4}+\frac{24}{6}+\frac{24}{6}=38 $$ I'm going to use this example to explain my formula above. We want a permutation of length $l$, so it's like having 4 slots to fill up with what we have. In this case we have our multiset $S=(3,2,1)$. We start with the first element $m_1=3$. How many of $m_1$ we can use, remembering that we have $l$ free slots and that after $m_1$ we have $m_2+m_3=3$ to use? The answer is $$ MAX(0,l-m_2-m_3) \leq x_1 \leq MIN(l,m_1) $$ We can't use less or more, or we won't reach $l$ at the end. So $x_1=(1,2,3)$. Let's start with $x_1=1$. What is the number of combinations with $4$ free slots and $1$ to use? This is the binomial coefficient $$ \binom{l}{x_1}=\binom{4}{1} $$ What I'm going to do, is to repeat what I did until I completely fill all the slots. Now we have $l-x_1=3$ free slots and $m_2$ to use. Of $m_2$ I can use no less than $MAX(0,l-x_1-m_3)$, and no more than $MIN(l-x_1,m_2)$. So $x_2=(2)$ as it's the only choice. What is the number of combinations with 3 free slots and 2 to use? Like before, it's $$ \binom{l-x_1}{x_2}=\binom{3}{2} $$ Combined with $x_1=1$ it becomes $$ \binom{l}{x_1}\binom{l-x_1}{x_2}=\binom{4}{1}\binom{3}{2} $$ Now we have $l-x_1-x_2=1$ free slots, and $m_3$ to use. One slot can be filled in only one way, so $x_3=1$ and the combinations are $$ \binom{l-x_1-x_2}{x_3}=\binom{1}{1} $$ Combined with $x_1=1$ and $x_2=2$ it becomes $$ \binom{l}{x_1}\binom{l-x_1}{x_2}\binom{l-x_1-x_2}{x_3}=\binom{4}{1}\binom{3}{2}\binom{1}{1} $$ We can clearly see a pattern here, so we can try to get a nicer formula from that $$ \binom{l}{x_1}\binom{l-x_1}{x_2}\binom{l-x_1-x_2}{x_3}=\frac{l!}{x_1!(l-x_1)!}\frac{(l-x_1)!}{x_2!(l-x_1-x_2)!}\frac{(l-x_1-x_2)!}{x_3!(l-x_1-x_2-x_3)!} $$ We can simplify that and get $$ \frac{l!}{x_1!x_2!x_3!(l-x_1-x_2-x_3)!} $$ Since we know that $x_1+x_2+x_3=l$ we can simplify again and get $$ \frac{l!}{x_1!x_2!x_3!(l-x_1-x_2-x_3)!}=\frac{l!}{x_1!x_2!x_3!0!}=\frac{l!}{x_1!x_2!x_3!} $$ $$ \frac{l!}{x_1!x_2!x_3!}=\frac{4!}{1!2!1!} $$ Now let's get back where we were. Since we filled up all the slots, we have to go back. Since $x_1=(1)$ and $x_2=(2)$, we have to get back to $x_1=(1,2,3)$ and pick the second, so $x_1=2$. $x_1=2$ implies $x_2=(1,2)$. Since there is more than 1 way to choose $x_2$, we can write it this way $$ \binom{4}{2}\left[\binom{2}{1}+\binom{2}{2}\right] $$ If we continue with $x_2=1$, we get $x_3=1$. And if we continue with $x_2=2$, we get $x_3=0$, and get $$ \binom{4}{2}\left[\binom{2}{1}\binom{1}{1}+\binom{2}{2}\binom{0}{0}\right]=\binom{4}{2}\binom{2}{1}\binom{1}{1}+\binom{4}{2}\binom{2}{2}\binom{0}{0}=\frac{4!}{2!1!1!}+\frac{4!}{2!2!0!} $$ Let's go back the last time to $x_1$ and pick up the last, so $x_1=3$. $x_1=3$ implies $x_2=(0,1)$, so $$ \binom{4}{3}\left[\binom{1}{0}+\binom{1}{1}\right] $$ If we continue with $x_2=0$, we get $x_3=1$. And if we continue with $x_2=1$, we get $x_3=0$ and get $$ \binom{4}{3}\left[\binom{1}{0}\binom{1}{1}+\binom{1}{1}\binom{0}{0}\right]=\binom{4}{3}\binom{1}{0}\binom{1}{1}+\binom{4}{3}\binom{1}{1}\binom{0}{0}=\frac{4!}{3!0!1!}+\frac{4!}{3!1!0!} $$ If we sum up all the results, we get the same formula as before.