I'm trying to prove that $ (n+1)!-1 = \sum_{i=1}^n i • i! $ combinatorially

Asked Dec 04 '23 at 14:42

Active Dec 04 '23 at 14:42

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I'm trying to prove this identity combinatorially:

$$ (n+1)!-1 = \sum_{i=1}^n i • i! $$

I tried to ask and answer the question: In how many ways can we list (n+1) different numbers such that we don't list the numbers in ascending order.

The left side is clearly (n+1)! -1, but how do I prove that that applies to the right side as well? I would appreciate any hints!

asked Dec 04 '23 at 14:42

Grant

Note: many of the arguments given in the duplicate are, in fact, proofs by induction, but there is a good combinatorial proof in the solutions. – lulu Dec 04 '23 at 14:49
@lulu Yes you are right, I really tried to find it before posting this question, but no result... Thank you – Grant Dec 04 '23 at 14:55

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