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Question


n=1C(n,0)+C(n,1)+...+C(n,n)P(n,n)==ek+m.Find k+m ?

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Solution

We have
n=1C(n,0)+C(n,1)+...+C(n,n)P(n,n)

=n=1nC0+nC1+...+nCnn!

=n=12nn!

=21!+222!+233!+...

=e21


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