Derangement is defined here on wikipedia as a permutation without fixed points. Consider the following generalization: an n-derangment of an m-set is a an n by m matrix in which each cell is a number from 1 to m and each column and row contains a number at most once. Two such matrices are equivalent if one can be created from the other by permuting columns. Notice in this definition we recover the normal notion of a derangement when we let n=2.
Recall that for large m, the proportion of arrangements that are derangements is about 1/e. I would like to know the analogous proportion for the generalized n-derangement. For the case of n=3, I suspect it can be calculated as follows however my "proof" is not really a proof but rather an intuition.
First pick the first row of the 3 by m matrix. Next pick the second row. There is a 1/e chance that its a derangement wrt the first row. Now pick the third row. There is a 1/e chance it is a derangement wrt the first and a 1/e chance it is a derangement wrt the second. So multiplying all these probabilities together we have a 1/e^3 chance that the entire thing is a 3-derangment. More generally I suspect the probability that an n-arrangment is an n-derangement is 1/e^(n choose 2).
I would like to know if I am wrong and if so what the correct answer is. Regardless if I am wrong I would like a proof of whatever the correct formula is. Thanks
PS:
Thanks to @bof for pointing out this is the same as the definition of latin rectangles. With this is mind, I am still interested in the answer to my question.
Your suspiciouns are correct. Let $L(n,k)$ be the number of $k\times n$ Latin rectangles, and let $D(n,k)$ be the number of $k$-derangements on $n$ symbols, so that $ D(n,k) = L(n,k)/n! $. It was first shown in $[1]$ that $$ L(n,k) \sim (n!)^k\exp\left(-\binom{k}2\right) $$ with $k$ fixed as $n\to\infty$, and which still holds as long as $k\in O((\log n)^{3/2-\epsilon})$. More precise estimates are provided in $[2]$, which hold for $k\in o(n^{6/7})$.
$[1]$ P. Erdös and I. Kaplansky, The asymptotic number of Latin rectangles, Amer. J. Math. 68 (1946), 230-236.
$[2]$ C.D. Godsil, B.D. McKay, Asymptotic enumeration of Latin rectangles, J. of Comb. Theory, Series B, 48, 1 (1990), 19-44, https://www.sciencedirect.com/science/article/pii/009589569090128M