4 votes 4 votes Consider a set $A$ with $6$ elements. Let $N_1$ denote the number of bijective functions from $A$ to $A$ and let $N_2$ denote the number of onto (surjective) functions from $A$ to $A.$ $N_2 - N_1 = \_\_\_\_\_.$ Combinatory go2025-dm-2 numerical-answers counting combinatory + – gatecse asked Jun 28, 2020 • edited Jul 23, 2020 by Lakshman Bhaiya gatecse 299 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
8 votes 8 votes For any finite set $A,$ a function from $A$ to $A$ is bijective if and only if it is injective if and only if it is surjective. So, the total number of bijective and surjective functions must be the same $(n!)$ and their difference equals $0.$ gatecse answered Jun 28, 2020 gatecse comment Share Follow See 1 comment See all 1 1 comment reply Pratik2404 commented Dec 11, 2023 reply Follow Share Sir, Can you please explain how is number of onto function is n! ? As per the question, it’s not mentioned for N1 that it’s bijective. Bijective condition is given for N1 only. 0 votes 0 votes Please log in or register to add a comment.
