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 $N^K - N$.