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

Akash Kumar Dutta , 12 Years ago
Grade 11
anser 1 Answers
Arun Kumar

Last Activity: 10 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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