# Calculating $\sum_{k=1}^nk(k!)$ combinatorially [duplicate]

The sum $\sum_{k=1}^nk(k!)$ can be easily calculated by noting $k(k!)=(k+1)!-k!$. Is there a way to calculate the sum nicely using a combinatorial argument. Is it possible to notice it is $(n+1)!-1$ combinatorially?

## marked as duplicate by Austin Mohr, Community♦Oct 11 '16 at 19:22

• @AlexR: I think what he meant was $(n+1)! - 1$, edited accordingly. – J. J. Sep 12 '14 at 11:57
• @J.J. Ah I figured so after the edit but decided to wait for comment. – AlexR Sep 12 '14 at 12:00
• I meant that each term was going to be $(k+1)!-k!$ – Jorge Fernández Hidalgo Sep 13 '14 at 15:18

I will explain it in case $n=4$ then I will replace it by general case. Assume you want to write a 5 digit number by numbers 1,2,3,4,5 except 12345. The number of possible numbers is 5!-1. In fact writing a 5 digit number by 1,2,3,4,5 has 5! possibilities and we delete 1 case.
The general case is same when you are counting the number of n+1 digits numbers using n+1 (ordered) alphabet except $\overline{123...n(n+1)}$.