Timeline for How to compute the number of combinations with r allowed repetitions [closed]
Current License: CC BY-SA 4.0
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| Nov 21, 2020 at 17:33 | history | closed |
Max Alekseyev coudy Desiderius Severus Amir Sagiv András Bátkai |
Not suitable for this site | |
| Nov 6, 2020 at 13:27 | review | Close votes | |||
| Nov 21, 2020 at 17:33 | |||||
| Nov 6, 2020 at 12:25 | comment | added | Mark Wildon | Of course. Thank you. | |
| Nov 6, 2020 at 11:48 | comment | added | Brendan McKay | @MarkWildon I/E isn't needed. The choices with some item chosen $r$ times equals the number with all items chosen at most $r$ times minus the number with all items chosen at most $r-1$ times. | |
| Nov 6, 2020 at 10:50 | comment | added | Mark Wildon | The generating function for the number $a_k$ of ways to choose $k$ items from $n$ with at most $r$ choices of each item is $(1+x+\cdots + x^r)^n$. Taking $r=1$ gives $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$ and letting $r$ tend to infinity one gets $1/(1-x)^n = \sum_{k=0}^\infty \binom{-n}{k}(-x)^k = \sum_{k=0}^\infty \binom{k+n-1}{k}x^k$, the two formulae in the question. As far as I know there are no very convenient formulae for $r$ in between. And then one would need some form of inclusion/exclusion to count those choices where some item is chosen exactly $r$ times. | |
| Nov 6, 2020 at 10:49 | history | edited | Mark Wildon | CC BY-SA 4.0 |
Tidied up formatting
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| Nov 6, 2020 at 10:28 | history | edited | YCor |
edited tags
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| Nov 6, 2020 at 10:26 | history | edited | Jara M | CC BY-SA 4.0 |
edited body
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| Nov 6, 2020 at 10:18 | review | First posts | |||
| Nov 6, 2020 at 11:05 | |||||
| Nov 6, 2020 at 10:16 | history | asked | Jara M | CC BY-SA 4.0 |