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Let $n\geq 0$ be an integer and let $0 \leq j \leq n$, $0 \leq t \leq n$.

Let $S_n(j, t)$ be the set of all permutations $\sigma$ of $\{1, \dots, n\}$ with the property that $$\sigma(i) \leq j \quad \text{for} \quad i \leq t.$$

Is there a simple formula for the cardinality of $S_n(j, t)$?

Example If $t>j$ then $S_n(j,t)=0$ by the pigeonhole principle ($\{1, \dots, t\}$ cannot map injectively into $\{1, \dots, j\}$ ).

Example If $t=j$ then the datum of an element of $S_n(j, t)$ is equivalent to the datum of a permutation of $\{1, \dots, j\}$ and a permutation of its complement in $\{1, \dots, n\}$. Hence $\# S_n(j,j) = j!(n-j)!$.

What about the general case $t\leq j$?

This question arose in a multilinear albgera calculation.

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Normally the cardinality of a permutation is $n! = \prod_{k=1}^n k$ by stating the first element has $n$ choices, the next $n-1$, etc.

Here we see that the first element has $j$ choices, the next $j-1$, etc, until $j + 1 - t$ after which the next element has no further restrictions so it has $n - t$ choices, then $n - t - 1$, etc, so we get $\prod_{k=1}^t (j + 1 - k) \times \prod_{k=t+1}^n (n + 1-k)$ as our answer.

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  • $\begingroup$ You're absolutely right, I think it's that simple! (I think the second product should start with $n-t$ rather than $n-t-1$, no?) $\endgroup$ – Bruno Joyal Sep 12 '18 at 16:41
  • $\begingroup$ @BrunoJoyal Fixed. $\endgroup$ – orlp Sep 12 '18 at 16:43
  • $\begingroup$ @orip Thanks :) $\endgroup$ – Bruno Joyal Sep 12 '18 at 16:45
  • $\begingroup$ @BrunoJoyal The reason for the error is that as a programmer I often count starting from zero so I thought that the interval with boundary $\leq t$ contained $t + 1$ elements but I forgot that it started at $1$, rather than $0$. $\endgroup$ – orlp Sep 12 '18 at 16:48

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