$$n!=\sum_{i=0}^n\binom ni(!(n-i))$$ Hi. I am trying to review Discrete Math 1 for the new term. I couldn't find a proper strategy to prove this statement. What would be the best way to prove such a question using equality?
2026-04-12 02:00:47.1775959247
Finding Number Of Derangement
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The number of arrangements of $(1,\dots,n)$ such that exactly $i$ elements are not moved equals $$\binom{n}{i}(!(n-i))$$
A summation of these terms gives all $n!$ arrangements.