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.
Normally the number of permutations of $\left\{1,2,\ldots,n\right\}$ 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.