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 May11 awarded Announcer Jan5 awarded Supporter Jan5 accepted Conditional probability exercise Jan5 comment Conditional probability exercise Thank you, I assumed I am looking for Pr(P|N) which was the wrong way. Jan5 asked Conditional probability exercise Oct26 awarded Editor Oct26 revised Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ edited body Oct26 awarded Student Oct26 awarded Custodian Oct26 reviewed Approve Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ Oct26 awarded Scholar Oct26 accepted Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ Oct26 comment Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ I see! It is 4 * (4^3 - 3^3). 4^3 is number of ways how to map remaining els MINUS the ways how to map remaining elements without mapping to the one element of B. Oct26 asked Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$