41 41 votes The minterm expansion of $f(P,Q,R) = PQ +Q \bar{R}+P\bar{R}$ is $m_2+m_4+m_6+m_7$ $m_0+m_1+m_3+m_5$ $m_0+m_1+m_6+m_7$ $m_2+m_3+m_4+m_5$ Digital Logic gatecse-2010 digital-logic canonical-normal-form normal + – go_editor 12.6k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments advaith_xyz commented Jul 8, 2024 reply Follow flag PQ we can write it as PQR' + PQR is minterm 6 and minterm 7 respectively so B and D are wrong P'Q'R' will not be present hence minterm 0 cannot be present so C cannot be correct answer. So the leftover option is A. 0 0 replyShare GO Classes commented Jun 24 i edited by Deepak Poonia Jun 24 reply Follow flag Watch the Detailed Video Solution by clicking the button below...!Watch Detailed Video Solution 2 2 replyShare Raj_Dev_Verma commented Aug 12 reply Follow flag m2+m4+m6+m7 option A is correct 0 0 replyShare Please log in or register to add a comment.
Best answer 54 54 votes $PQ + QR' + PR' = PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R' $ $\quad= PQR + PQR' + P'QR' + PQ'R' (111 + 110 + 010 + 100)$ $\quad =m_7 + m_6 + m_2 + m_4$ Option A. Alternatively, Using K-map Arjun answered Nov 7, 2014 • edited Apr 24, 2019 by ajaysoni1924 Arjun comment Share Follow See all 6 Comments 6 6 Comments reply Show 3 previous comments Lakshman Bhaiya commented Oct 13, 2018 reply Follow flag @raushan sah You can expand all the term $f(P, Q, R)=PQ+Q\bar{R}+P\bar{R}$ $f(P, Q, R)=PQ(R+\bar{R})+(P+\bar{P})Q\bar{R}+P(Q+\bar{Q})\bar{R}$ $f(P, Q, R)=PQR+PQ\bar{R}+PQ\bar{R}+\bar{P}Q\bar{R}+PQ\bar{R}+P\bar{Q}\bar{R}$ $f(P, Q, R)=PQR+PQ\bar{R}+\bar{P}Q\bar{R}+P\bar{Q}\bar{R}$ [Remove the similar term] In case of Sum of Product,we all know variable represent value $'1'$ and complement of variable represent the value $'0'.$ $f(P, Q, R)=PQR(111)+PQ\bar{R}(110)+\bar{P}Q\bar{R}(010)+P\bar{Q}\bar{R}(100)$ We can also represent,these minterm as $f(P, Q, R)=\sum(7,6,4,2)$ and $m_{2}+m_{4}+m_{6}+m_{7}$ is also correct. [Because we write sum of min-term=sum of product term(Canonical)] 6 6 replyShare raushan sah commented Oct 13, 2018 reply Follow flag @lakshman thanx yes got it 1 1 replyShare talha hashim commented Nov 5, 2018 reply Follow flag nice @arjun sir 0 0 replyShare Please log in or register to add a comment.
5 5 votes option A =PQ(R+~R)+(P+~P)Q~R+P~R(Q+~Q) on simplification we will get minterms as 2,4,6,7 saipriyab answered Nov 5, 2017 saipriyab comment Share Follow 0 reply Please log in or register to add a comment.
3 3 votes Answer: OPTION A vannjot answered Sep 23, 2019 vannjot comment Share Follow 0 reply Please log in or register to add a comment.