Statement I: n n − ( 1 n ) ( n − 1 ) n + ( 2 n ) ( n − 2 ) n − ⋯ + ( − 1 ) n − 1 ( n − 1 n ) = n ! . Statement II: If two finite sets have the same number of elements, the numbers of onto and on

Question

Statement I: n n − ( 1 n ​ ) ( n − 1 ) n + ( 2 n ​ ) ( n − 2 ) n − ⋯ + ( − 1 ) n − 1 ( n − 1 n ​ ) = n ! . Statement II: If two finite sets have the same number of elements, the numbers of onto and one-to-one functions between them are equal.

TestHub · Accepted Answer

Inclusion-exclusion counts the onto maps from an n -element set to an n -element set by the expression in I. Such maps are bijections, counted by n ! , exactly as II states.

Mathematics - Permutation & Combination Question with Solution | TestHub

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