0

Guest

Courses New

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
8 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

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More