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seen Jul 3 at 11:16

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 suggested edit on 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 $f(1) > f(2) > f(3)$