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Say there are $n$ elements where some of the $n$ elements are repeated elements. For example, there are $10$ dogs and there are $2$ dogs of species $A$, $3$ of species $B$ and $5$ of species $C$.

Now how many ways are there to permute $k$ elements from the $n$ elements?

For example, how many ways are there to select $4$ dogs from the $10$ dogs (order matters).

I know that for $k = 10$, the answer is

$$\frac{10!}{2!\cdot3!\cdot5!}$$

But what about general $k \neq n$?

I thought maybe answer would be

$$\frac{nP_k}{n_1!n_2!n_3!...}$$

but it doesn’t seem right?

What is answer with full explanation please, thank you. Actually even hints are fine. I am trying to make my concepts more clear basically.

Note: I am looking for a general formula, not a strategy to solve. I can solve these kind of questions by considering cases but I want a general formula

N. F. Taussig
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  • Please if any detail is not clear let me know. i will fix it –  Jun 27 '22 at 19:55
  • See also https://math.stackexchange.com/questions/4011275/write-five-digits-numbers-using-0-0-0-1-1-2-2-2-2-3-3-3-4-4/ for a method to compute these numbers. There is no general formula, only computational methods. – Mike Earnest Jun 28 '22 at 17:41

2 Answers2

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It depends very much on the multiplicities of the elements. Suppose your multiset is

$$ \{\underbrace{1,\cdots,1}_{m_1},\underbrace{2,\cdots,2}_{m_2},\cdots\} $$

with $m_1+m_2+\cdots=n$ elements. If you pick $k$ out of the $n$ elements, and more specifically you pick $k_1$ ones, $k_2$ twos, $k_3$ threes and so on, with $k=k_1+k_2+\cdots$ then the number of ways to permute the $k$ elements is the multinomial coefficient $\binom{k}{k_1,k_2,\cdots}$ by a standard argument.

Tallying this over all possible $k_i$s, we get

$$ \sum_{\substack{0\le k_i\le m_i \\ k_1+k_2+\cdots=k}} \binom{k}{k_1,k_2,\cdots} $$

I expect this will not simplify.

anon
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  • this is related to generating functions? –  Jun 27 '22 at 20:09
  • @gla Almost anything in combinatorics is "related to generating functions," even if sometimes you have to squint hard enough. Gonna have to be more specific. – anon Jun 27 '22 at 20:12
  • because you said coefficient I thought it was related. Because actually i have not learned generating functions but I am going to learn it now. is the "standard argument" a generating functions proof? –  Jun 27 '22 at 20:14
  • @gla Ah, I didn't realize you were talking about multinomial coefficients in particular. Yes, $\binom{k}{k_1,k_2,\cdots}$ is the coefficient of $x_1^{k_1}x_2^{k_2}\cdots$ when you expand out $(x_1+x_2+\cdots)^k$. The standard argument for multiset permutations being multinomial coefficients is essentially the orbit-stabilizer theorem, though elementary discussions will rarely call this out because abstract algebra will seem to most authors to be too advanced for their target audience. You can google "multiset permutations" or "permutation MISSISSIPPI" for a taste. – anon Jun 27 '22 at 20:22
  • Although I guess you could provide a generatingfunctionological explanation as well, since the terms you get when you expand $(x_1+x_2+\cdots)^k$ (assuming you maintain order of variables in the monomials and do not combine like terms) essentially correspond to $k$-permutations of ${1,2,\cdots}$. – anon Jun 27 '22 at 20:23
  • ok unfortunately i don’t have enough reputation to upvote u. i won't accept the answer in the hopes of a combinatorial proof building from basic principles, but if i don’t get one in some time, i will accept yours. thank you once again for answering my questions. –  Jun 27 '22 at 20:25
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If you want a general formula rather than a case by case, you'll have to use generating functions, which here can be understood quite simply.

For each type of dog, we shall use $x$ as a sort of place holder with the coefficient of x denoting the number taken from each type, taking care to divide by the appropriate factorial when more than $1$ of that type is taken, e.g.
$(x^0 + x^1 + x^2/2! + x^3/3! +x^4/4!+x^5/5!)$, would select $0,1,2,3,4 \;or\; 5$ from species $C$ dogs.

So what we do is to find $4!$ times the coefficient of $x^4$ considering each species of dog in $(x^0+ x^1+x^2)(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3+x^4+x^5)$

which works out to $264$

true blue anil
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  • i will come back to this once i have thoroughly learnt generating functions; thank you –  Jun 27 '22 at 20:37
  • Btw, I wrote out formulas in full for better understanding. We could have written x^0 as 1, x^1 as x and left out x^5 in the last term as we aren't selecting more than 4 dogs. – true blue anil Jun 27 '22 at 20:40