Computing Permutations with Partial Duplicates - MathOverflow

Computing Permutations with Partial Duplicates

2011-09-14T20:59:21Z

I am looking for a way to compute the number of $K$ permutations of a multiset with $N*D$ elements where each group has exactly $D$ equal elements (and typically $D < N$ ).

I've got an application that actually generates these unique permutations and works on them, but I'd like to understand how I can compute the number of sets I'll have across various inputs without computing the entire result.

Example (in R):

N <- 19
K <- 4
# Implied D = 3 by just duplicating it in-place three times.

a <- append(1:N, append(1:N, 1:N))
b <- unique(gtools::permutations(length(a), K, a, set=FALSE))

nrow(b) in this case will be 130,302.

This is slow and inelegant. Can someone help me do this with actual math?

Expanding a bit

If N is 9 and D is 3, my input might look like this:

1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

A standard permutation would look like this:

1    1    1    2
1    1    1    2
1    1    1    2
1    1    1    3
1    1    1    3
1    1    1    3

But at this point, I want to treat the things that look the same as the same, so I deduplicate to get the following:

1    1    1    2
1    1    1    3
1    1    1    4
1    1    1    5
1    1    1    6
1    1    1    7

The first (full permutation) provides 421,200 rows: $(9*3)! \over (9 * 3 - 4)!$

My final, deduplicated answer is 6,552 rows. I'd like to know how I can get that without generating them all.

New Discovery

For my initial case where $D = K - 1$, I get the correct answer with $N^K - N$.

Answer by psd for Computing Permutations with Partial Duplicates

2011-09-14T22:55:24Z

oeis...........

Answer by nullghost for Computing Permutations with Partial Duplicates

2011-09-14T23:04:38Z

I'm thinking the result should be:
n ^ k for d >= k
n ^ k - n ^ (k - d) for d < k

Answer by Max Alekseyev for Computing Permutations with Partial Duplicates

2011-09-15T09:12:48Z

The number of $K$-permutations of the numtiset ${ 1^D, 2^D, \ldots, N^D }$ is $$\sum_{j_1+j+2+\dots+j_N=K\atop 0 \leq j_i \leq D} \binom{K}{j_1,j_2,\dots,j_N}.$$ (summands here are multinomial coefficients)

Alternatively, denoting by $m_i$ the number of $j$'s equal $i$, we get a formula as the sum over restricted partitions: $$\sum_{0m_0+1m_1 + \dots + Dm_D = K\atop m_0 + m_1 + \dots + m_D = N,\quad m_i\geq 0} \binom{N}{m_0,\dots,m_D} \frac{K!}{1!^{m_1} 2!^{m_2} \cdots D!^{m_D}}$$

For the example with $N=9$, $D=3$, $K=4$, the latter formula consists of four summands and gives: $$\binom{9}{5,4} \frac{4!}{1!^4} + \binom{9}{6,2,1}\frac{4!}{1!^2 2!^1} + \binom{9}{7,1,1}\frac{4!}{1!^1 3!^1} + \binom{9}{7,2}\frac{4!}{2!^2}$$ $$ = 3024 + 3024 + 288 + 216 = 6552$$ as expected.

Answer by nomatter for Computing Permutations with Partial Duplicates

2011-09-15T11:09:00Z

You could try to use exponential generating functions.

For each of the N letters you could use the exponential generating function

$$ \sum_{i=0}^D \frac{x^i}{i!} $$

Cause each letter can be used at most D times this is the same for each letter.

Then for using all different letters the egf's have to be multipled (you have N different letters so N times):

$$ \left(\sum_{i=0}^D \frac{x^i}{i!}\right)^N $$

Then you are looking for the amount of different words of length K which is if you expand the expression above (which is possible if you put into it values for $D, N$. Then the coefficient of $x^K$ multiplied by $k!$ is your solution.

What I wasn't able to do right now is trying to expand the generating function into a series without using concrete values for $D$ and $N$.