TITLE: Calculating $\sum_{k=1}^nk(k!)$ combinatorially
QUESTION [7 upvotes]: 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?

REPLY [7 votes]: 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.
Now count it in this way;

Some of them can be written with 5 not be 5th one. So for 5 we have 4 positions and the other 4 numbers should be put in 4 places by 4! possibilities so 4(4!)
The other cases 5 should remain at 5th position so again divide to two cases. First, 4 not in 4th position you so 4 has to be in 1st, 2nd or 3rd position by 3 choices then 1,2,3 have to be put at 2 remained positions from 1st, 2nd and 3rd which 4 is not there plus 4th position which is empty by 3! ways so this be done by 3(3!).
The remained cases are ones that 5 is at 5th and 4 is at 4th position. Again divide it into two cases. First, 3 is not at 3rd position so we have 2 choice for it, 1st or 2nd position. 1 and 2 now can be sit at the remained position from 1st and 2nd which 3 is not there plus 3rd position so they can be put in the remained possible places by 2!. So number of numbers which produced here are 2(2!).
Now the at remained possibilities, 5 is at 5th, 4 is at 4th, 3 is at 3rd positions. As we don't want 12345. So 2 should be at 1st (by 1 choice) and 1 should be at 2nd (by 1!=1 choice) so 1(1!).

By the second method we count possible requested numbers as this sum 4(4!)+3(3!)+2(2!)+1(1!). And as we were counting same thing so we have to have (4+1)!-1=5!-1=4(4!)+3(3!)+2(2!)+1(1!).
The general case is same when you are counting the number of n+1 digits numbers using n+1 (ordered) alphabet except $\overline{123...n(n+1)}$.