0 votes 0 votes Digital Logic combinatory boolean-algebra + – Pradip Nichite asked Jan 24, 2016 • retagged Jun 25, 2017 by Arjun Pradip Nichite 5.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes For number of binary operation (lets denote set with R) since |R|=4 binary operation( R X R ->R) would be 4^(4*4) =4^16 Abhishekcs10 answered Jan 24, 2016 • edited Jan 24, 2016 by Abhishekcs10 Abhishekcs10 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes No of binary operation on a set with n element=n^(n^2) here n=4 Ans 4^16 ManojK answered Apr 18, 2016 ManojK comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes For number of binary operation with n variable are 22^n Here Cardinality of R is 4. No of Function = 22^4 = 216 Digvijay Pandey answered Apr 19, 2016 Digvijay Pandey comment Share Follow See 1 comment See all 1 1 comment reply MIRIYALA JEEVAN KUMA commented Jan 11, 2018 reply Follow Share you wrote the formula for the number of binary functions with n binary variables. it's not correct. please check. 0 votes 0 votes Please log in or register to add a comment.