In combinatorics, there is a formula "$n$ multichoose $k$", which is the way of making a multiset having $k$ elements choosing out of $n$ options. "$n$ multichoose $k$" is the same as "$(n+k-1)$ choose $k$". Why is that?
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Just to clarify, are you asking in how many $k$ objects can be selected from a multiset containing $n$ different types of objects? – N. F. Taussig Jun 26 '15 at 21:18
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See http://mathworld.wolfram.com/Multichoose.html – Amy B Jun 26 '15 at 21:25
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1See Theorem Two here; the explanation is quite clear. – Brian M. Scott Jun 26 '15 at 21:28
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@Taussig: what I have in mind is wolfram's example, someone who wants 5 pinches out of 9 spices – Jun 26 '15 at 22:00
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Somehow I explained where the formula comes from in an answer to this question. It is not a proof but rather the intuition. – Carlos H. Mendoza-Cardenas Nov 25 '15 at 12:56
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There is an excellent tabular illustration in Knuth. TAOCP Volume 4A (Combinatorial Algorithms), section 7.2.1.3, Table 1 and its surrounding discussion do a superb job of teaching intuitively how a multichoose directly corresponds to a primal combination. – inopinatus Jun 07 '23 at 00:23
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Let's say you have $N$ items (all alike for now) and $K-1$ vertical bars (all alike for now).
How many unique ways can you line up the $N$ items and the $K-1$ vertical bars?
Now pretend that everything to the left of the first bar is Type $1$, everything between the first and second bar is Type $2$ ... and everything to the right of the last $(K-1)$th bar is Type $N$.
Do you see the connection?
John
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I have seen this argument before, but it still isn't clear to me. I am learning from Ardila [link]https://www.youtube.com/watch?v=vKskmKq3v40&list=PL-XzhVrXIVeSi7xym1XAfFIxOAaHVhjtP&index=3[/link] and he gives a proof from 5:30 from 7:30. Claiming a bijection between multichoose and weak compositions. Nevertheless, this bijection is not very clear to me – Jun 26 '15 at 22:03