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how to derive the formula for number of surjections possible in a relation? formula being:- n sigma (-1)^(n-r).nCr.(r)^n r=0

how to derive the formula for number of surjections possible in a relation?
formula being:-
n
sigma (-1)^(n-r).nCr.(r)^n
r=0

Grade:11

1 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
7 years ago
Hi
See you're not able to express your doubt but anyways probably by following formulae.

 \sum_{j=0}^n (-1)^j\tbinom n j P(n-j) = n!a_n

 \sum_{j=0}^n (-1)^j\tbinom n j P(m+(n-j)d) = d^n n! a_n

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

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