30 votes 30 votes The operation which is commutative but not associative is: AND OR EX-OR NAND Digital Logic gate1992 easy digital-logic boolean-algebra multiple-selects + – Kathleen asked Sep 12, 2014 • edited Apr 19, 2021 by Lakshman Bhaiya Kathleen 7.4k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply biranchi commented Jul 8, 2016 reply Follow Share 'communicative' should be 'commutative' 3 votes 3 votes Deepak Poonia commented Jul 16, 2022 i edited by Deepak Poonia Sep 25, 2023 reply Follow Share $\color{red}{\text{Find Detailed Video Solution Below}}$ $\color{BLACK}{\text{ , with Complete Analysis:}}$ https://www.youtube.com/watch?v=ODeuXLf7rzU&list=PLIPZ2_p3RNHi0XVjX565f6vRfT1nRS-X0&index=6 1 votes 1 votes mohit7891 commented Nov 6, 2023 reply Follow Share @GO Classes I always confused between properties of commutative vs associative, can anyone help me? 0 votes 0 votes Deepak Poonia commented Nov 6, 2023 reply Follow Share @mohit7891 , Refer the following lectures: 1. Associative, Commutative Property 2. Questions on Associative, Commutative Property 1 votes 1 votes Please log in or register to add a comment.
Best answer 43 votes 43 votes The answer is D. Remark: Every logic gate follows Commutative law. AND, OR, Ex-OR, EX-NOR follows Associative law also. NAND, NOR doesn’t follow Associative law. ankitrokdeonsns answered Oct 20, 2014 • edited Apr 19, 2021 by Lakshman Bhaiya ankitrokdeonsns comment Share Follow See all 6 Comments See all 6 6 Comments reply Sachin Mittal 1 commented Dec 25, 2016 reply Follow Share Its good to remember- 1)Every logic gate follows Commutative law. 2)AND,OR,Ex-OR,EX-NOR follows Associative law also. NAND,NOR doesn`t follow Associative law. 34 votes 34 votes Vidhya Pate commented Mar 21, 2017 reply Follow Share commutative law: A operator B = B operator A associative law: A operator (B operator C) = (A operator B) operator C 8 votes 8 votes Sumaiya23 commented Jan 19, 2018 reply Follow Share EX-NOR is associative... 2 votes 2 votes chauhansunil20th commented Nov 13, 2018 reply Follow Share Every logic gate follows Commutative law. how? A->B is not same as B->A, right? 0 votes 0 votes akash.dinkar12 commented Jan 7, 2019 reply Follow Share @chauhansunil20th The implication is a logic Gate?? 0 votes 0 votes chauhansunil20th commented Jan 7, 2019 reply Follow Share @akash.dinkar12 why not? why even this doubt? 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes Ans is D you can varify it like ((AB)'C)' = (A(BC)')' are not equivalent for input 011. Brij Mohan Gupta answered Apr 9, 2017 Brij Mohan Gupta comment Share Follow See 1 comment See all 1 1 comment reply Chandrabhan Vishwa 1 commented Dec 12, 2017 reply Follow Share yes option D 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes NAND gate is commutative but not associative. abhishekmehta4u answered May 24, 2018 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes We get for inputs 001,011,100 and 110 both the NAND and NOR are not associative sutanay3 answered May 24, 2018 sutanay3 comment Share Follow See all 0 reply Please log in or register to add a comment.