Hello,

This seems to involve Rencontres numbers $D_{n,r}$, the number of permutations in symmetric group $S_n$ with $r$ fixed points, for example, see OEIS Rencontres numbers or this post, The number of cycles in a random permutation, in the blog of Professor Tao.

Let $P_n$ be the set of partitions of $10$ into $n$ (not necessarily distinct and not necessarily non-zero) components. I.e,

$$ P_n = \{(a_1,a_2,\dots,a_n): \sum_{i}a_i = 10, 0 \leq a_i \leq 10\}. $$

Then the probability that $T=n$ is

$$ P(T=n) = \sum_{\pi = (a_i)\in P_n}\left(\frac{D_{10,a_1}}{10!}\right)\left(\frac{D_{10-a_1,a_2}}{(10-a_1)!}\right)\dots\left(\frac{D_{10-a_1-\dots-a_{n-1},a_n}}{(10-\sum_{i=1}^{n-1}a_i)!}\right). $$

Hello,

This seems to involve Rencontres numbers $D_{n,r}$, the number of permutations in symmetric group $S_n$ with $r$ fixed points, for example, see OEIS Rencontres numbers or this post, The number of cycles in a random permutation, in the blog of Professor Tao.

Let $P_n$ be the set of partitions of $10$ into $n$ (not necessarily distinct and not necessarily non-zero) components, where order matters. I.e,

$$ P_n = \{(a_1,a_2,\dots,a_n): \sum_{i}a_i = 10, 0 \leq a_i \leq 10\}. $$

Then the probability that $T=n$ is

$$ P(T=n) = \sum_{\pi = (a_i)\in P_n}\left(\frac{D_{10,a_1}}{10!}\right)\left(\frac{D_{10-a_1,a_2}}{(10-a_1)!}\right)\dots\left(\frac{D_{10-a_1-\dots-a_{n-1},a_n}}{(10-\sum_{i=1}^{n-1}a_i)!}\right). $$