How many permutations of {1,2,3,4,5} have no element in its original position? (For example, (2,1,4,5,3) counts but (2,1,3,5,4) does not since 3 is in position 3.)
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By inclusion-exclusion, the count is 5!−(15)4!+(25)3!−(35)2!+(45)1!−(55)0!=120−120+60−20+5−1=44.