A={1,2,...n}
B={1,2,...n}
total no. of functions from A-->B
each element of A has exactly n options (ie card(B) options).
so total no of functions possible = card(B)^card(A) =n^n
total no. of one one functions.
first element of A has n options in B to map with. 2nd element has n-1 options in B to map with and so on.
so. n(n-1)(n-2)...[n-(n-1)] = n!
required probability = n!/n^n =
(n-1)!/n^(n-1)option d
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"You don't have to believe in God, but you should believe in The Book." -Paul Erdős
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