MHB What is the relationship between r-multisets and combinations of 0s and 1s?

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The number of r-multisets with n distinct objects is equivalent to the number of ways to arrange r zeros and (n-1) ones in a row. This relationship arises because forming r-multisets involves creating tuples where the sum of object counts equals r, with each count being non-negative. Arranging the zeros and ones effectively divides the zeros into n groups, which corresponds to the multisets. This concept highlights the combinatorial connection between multisets and binary arrangements. Understanding this relationship clarifies the structure of r-multisets in combinatorial mathematics.
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The no.of r-multisets with n distinct objects is
equal to the number of ways of combining r-0s and (n-1) 1s.
How?Can anyone explain this?
 
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Re: The no:of r-multisets

mathmaniac said:
The no.of r-multisets with n distinct objects is
equal to the number of ways of combining r-0s and (n-1) 1s.
I assume that $r$-multisets means multisets of cardinality $r$, and by combining zeros and ones the problem means arranging them in a row. The number of multisets of cardinality $r$ consisting of $n$ distinct objects equals the number of tuples $(r_1,\dots,r_n)$ such that $r_i\ge0$ for all $i$ and $\sum_{i=1}^n r_i=r$. Here $r_i$ serves as the number of copies of object number $i$. In turn, arranging $r$ zeros in a row and inserting $n-1$ ones between them amounts to dividing zeros into $n$ groups, some of which may be empty. Thus, each arrangement gives rise to a tuple described above.

See also Theorem 2 on this Wiki page.
 
Re: The no:of r-multisets

Thanks for replying,E.M

but I had got it on my own.
 
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