Question 153
Let $p = p_{1} p_{2} \ldots p_{6}$ be a permutation of the integers from $1$ to $6$. For any such permutation $p$, we count how many integers there are which have nothing bigger on its left. We let $f(p)$ be the number of these integers for the permutation $p$. For example, $f(612345) = 1$ because only $6$ has nothing to its left which is bigger. On the other hand, $f(135462) = 4$ because only $1, 3, 5$, and $6$ satisfy the condition.
Let $S$ be the sum of $f(p)$ over all the $6!$ different permutations. Find the sum of the digits of $S$.
Let $S$ be the sum of $f(p)$ over all the $6!$ different permutations. Find the sum of the digits of $S$.