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 Feb 17 awarded Popular Question May 11 awarded Announcer Jan 5 awarded Supporter Jan 5 accepted Conditional probability exercise Jan 5 comment Conditional probability exercise Thank you, I assumed I am looking for Pr(P|N) which was the wrong way. Jan 5 asked Conditional probability exercise Oct 26 awarded Editor Oct 26 revised Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ edited body Oct 26 awarded Student Oct 26 awarded Custodian Oct 26 reviewed Approve Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ Oct 26 awarded Scholar Oct 26 accepted Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$ Oct 26 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. Oct 26 asked Number of surjective functions$f: A->B$ where $f(1) > f(2) > f(3)$