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4 Answers

Best answer
43 votes
43 votes

So, the equivalent expression will be $\bar AC + A\bar C + A\bar B$ 

(B) option

edited by
13 votes
13 votes
Let us solve the given function

$F = AB’C + A’BC + ABC’ + A’B’C +  AB’C'$

$F = AB'(C + C') + A'C(B + B') + ABC'$

$F = AB'.1 + A'C.1 + ABC'   [C + C' = 1,  B + B '= 1 ]$

$F = AB' + A'C + ABC'$

$F = A(B' + BC')+ A'C$

$F=A( ( B'+B ).( B' + C' ) ) + A'C \:\:\:\:\:[\because \text{Using Distributive law}: A + B.C = (A + B).(A + C)]$

$F = A.1.( B' + C' ) + A'C  \:\:\:\:\:\:\:\:[\because B' + B = B + B' = 1  ]$

$F = A.(B' + C' ) + A'C$

$F = AB' + AC' + A'C$
So, the correct answer is $(B)$.
edited by
3 votes
3 votes
AB’C + A’BC + ABC’ + A’B’C + AB’C’

AB’(C + C') + A'C ( B+ B') +  AC’ ( B + B' )

AB' + A'C + AC'

Hence option b ..!
Answer:

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