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If I have a set $\{1,2,...,m\}$, how do I find the cardinality of the set $\{\pi \in S_m : |{\rm supp}(\pi)|=k\}$ where $0 \leqslant k \leqslant m$.

I ask because I previously thought I found the answer. But all I had found was the number of $k$-cycles. I forgot to take into account that a permutation can be written as a product of distinct cycles.

I’m not sure how to calculate this now that $\pi$ could be the product of any number of distinct cycles.

Shaun
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Partey5
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  • Is the answer $mCk \cdot (k-1)!$ because the number of ways to choose $k$ from $m$ colours to permute is $mCk$ and the number of ways to permute $k$ items with an empty support is $(k-1)!$ from my own calculations. – Partey5 Mar 23 '20 at 23:13
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    Do you know how to compute the number of derangements? – saulspatz Mar 23 '20 at 23:22
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    I think the correct aswer is $C_m^k\cdot D_k$, because you have to choose $k$ elements between $m$ elements and compute the numbers of permutations such each element is not in the initial position. Thus, the answer will be $$\frac{m!}{(m-k)!}\left(\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^k}{k!}\right).$$ – DiegoMath Mar 23 '20 at 23:23
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    Look at https://math.stackexchange.com/questions/2465803/whats-the-probability-that-a-given-permutation-has-exactly-k-fixed-points – saulspatz Mar 23 '20 at 23:29
  • Thank you. I’ve never heard of dearrangements before. Why is what I said not correct. It seems logical to me. Is there some flaw? – Partey5 Mar 23 '20 at 23:37
  • Some thoughts: A $k$- cycle has support $k$, yes. There are also the products of disjoint cycles whose length adds up to $k$. This complicates it.

    The number of $k$-cycles is ${}_mP_k/k$.

    Now a product of disjoint cycles whose length adds up to $k$ corresponds to a partition of $k$.

     –  Mar 23 '20 at 23:38
  • So was my $(k-1)!$ statement incorrect. When I worked that out I drew 4 LHS dots and 4 RHS dots. The first dot has 3 options to go to as it can’t go to itself. If I then go to a dot that hasn’t been mapped onto yet then this dot has 2 options. I paid it up with one of them. Then let the 3rd dot be one that hasn’t been mapped onto itself yet. It has one option. Pair it up. Then pair the remaining one. – Partey5 Mar 23 '20 at 23:41
  • Maybe this doesn’t work because I had to work through the dots in a specific order to get my answer. – Partey5 Mar 23 '20 at 23:43
  • Thank you. I had never heard of derangement before and my problem is solved now. Thank you. – Partey5 Mar 23 '20 at 23:57

1 Answers1

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The correct aswer is $C_m^k\cdot D_k$, because you have to choose $k$ elements between $m$ elements and compute the numbers of permutations such each element is not in the initial position. Thus, the answer will be $$\frac{m!}{(m-k)!}\left(\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^k}{k!}\right).$$

DiegoMath
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