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Suppose n people leave their coats at the cloakroom, but on leaving the cloakroom the supervisor randomly gives any coat back to each person. Now to determine the number of permutations in which s persons receive their coat, there is the formula:

$$\sum\limits_{t=s}^{n}(-1)^{t-s}\binom{t}{s}\binom{n}{t}(n-t)!.$$

Now I would like to understand the formula. To do this, I will consider the case n=3 and s=1 to be able to illustrate this and determine the subset: N$_{1}$={Lists with 1 in position 1} (I mean: person 1 gets his coat back). N$_{2}$ = {Lists with 2 in position 2} and N$_{3}$ = {Lists with 3 in position 3}

Then

$$\sum\limits_{t=1}^{3}(-1)^{t-1}\binom{t}{1}\binom{3}{t}(3-t)! = \binom{1}{1}\binom{3}{1}(3-1)! - \binom{2}{1}\binom{3}{2}(3-2)! + \binom{3}{1}\binom{3}{3}(3-3)!.$$

Now consider a I multiplication: For example $\binom{1}{1}\binom{3}{1}(3-1)!$.

The part $\binom{3}{1}(3-1)!$ means: I think this is the same as |N$_{1}$|+|N$_{2}$|+|N$_{3}$| and that's mean we fix 1 of 3 that's why $\binom{3}{1}$ and the other 3-1 elements may be anywhere, that's mean (3-1)!

I know that the inclusion-exclusion principle applies here.

Now I try to understand the coefficient $\binom{1}{1}$ or in general, how can I imagine the coefficient $\binom{t}{s}$.

I am very grateful if you can explain this to me.

RobPratt
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1 Answers1

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$\binom{n}{t}\binom{t}{s}$ counts the number of pairs of set $(T,S)$ such that $S\subseteq T\subseteq \{1,2,\dots,n\}$ and $|S|=s,|T|=t.$

This can be shown by picking $T$ in $\binom{n}t$ ways, then picking $S$ from $T$ in $\binom{t}s$ ways.

So you can rewrite your formula as:

$$\sum_{|S|=s}\sum_{T\supseteq S} (-1)^{|T|-s}(n-|T|)!$$

Essentially, for fixed $S,$ $$\sum_{T\supseteq S}(-1)^{|T|-s}(n-|T|)!$$ is the number of ways that exactly the people in the set $S$ get their coats.

The term $(n-|T|)!$ counts all permutations that give back coats to at least the people in $T.$ Basically, permutations that are the identity on $T$ and any permutation of the $n-|T|$ remaining elements.

It is easier to explain this formula if you remember that:

$$\binom{n}{t}\binom ts=\binom ns\binom{n-t}{t-s}$$

We can see the right side is the number of ways to pick $S$ first, and the number of ways to pick $t-s$ additional elements to add to $S$ to get $T.$

Then your formula becomes, letting $r=t-s,$ $$\binom{n}s\sum_{r=0}^{n-s}(-1)^r\binom{n-s}{r}(n-s-r)!$$

Now the factor $\binom ns$ corresponds to picking just one set of $s$ people who we want to be exactly who gets their coats back, and the sum over $r$ is correcting for the cases when at least $|S|$ plus an additional $r$ got their coat back.

But in the original formula, $\binom nt$ corresponds to the number of ways to pick $T,$ and $\binom ts$ corresponds to the number of $S$ we’ve over-counted or under-counted in our inclusion-exclusion for permutations which fix $T.$

Thomas Andrews
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  • I'll be honest, I don't understand it. That's why I'm trying to understand the specific case for n=3 and s=1. –  Apr 08 '22 at 19:56
  • $\binom{3}{t}\binom{t}1$ counts the number of pairs of sets $S\subseteq T\subseteq {1,2,3}$ such that $|S|=1$ and $|T|=t.$ Then $(n-t)!$ counts how many permutations fix $T$ and possibly some more elements. You can think of $\binom 3t\binom t1$ as picking $t$ elements first, then picking $1.$ That one is the “one element” we want fixed, but we have to correct for counting in the cases where we fix more than one. – Thomas Andrews Apr 08 '22 at 20:14
  • if I consider, for example: $\binom{2}{1}\binom{3}{2}$ (3-2)! $\binom{2}{1}$ means to select exactly one property from the two intersections (i.e. that exactly one person gets their coat back). From these 3 people, one searches for the properties where exactly 2 persons get their coat back, i.e. the two intersections (this is $\binom{3}{2}$) the remaining people 3-2 are permuted, therefore (3-2)! I know it makes less sense for s=2. Have I understood the content correctly? –  Apr 08 '22 at 21:53
  • Do you see why $\binom{3}{t}\binom{t}1=\binom{3}1\binom{3-1}{t-1}?$ That might be a more profitable way to see it, because $\binom{3}{1}$ counts the number of sets that you want to fix, and $t-1$ is the number additional elements, for permutations that fix more than just our set of $s=1$ element. – Thomas Andrews Apr 08 '22 at 22:11
  • Can I replace $\binom{3}{t} \binom{t}{1}$ with $\binom{3}{1} \binom{3-1}{t-1}$? For t=1 the interpretation is: $\binom{3}{1}$ = counts the number where exactly one gets his coat back (I mean the sets N${1}$, N${2}$,N${3}$). $\binom{2}{1}$ = counts the number of where exactly one gets his coat back of two sets. And the remaining 3-2 are permuted, therefore (3-2)!. Did I understand this correctly? A second question. Can I interpret the $\binom{t}{s}$ as how often an element is counted with |N${1}$|+|N${2}$|+|N${3}$|. –  Apr 09 '22 at 10:23