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The simplified SOP (Sum of Product) from the Boolean expression

$$(P + \bar{Q} + \bar{R}) . (P + {Q} + R) . (P + Q +\bar{R})$$ is

1. $(\bar{P}.Q+\bar{R})$
2. $(P+{Q}.\bar{R})$
3. $({P}.\bar{Q}+R)$
4. $(P.Q+R)$

$(P + \bar{Q} + \bar{R}) . (P + {Q} + R) . (P + Q +\bar{R})$

$(P + \bar{Q} + \bar{R}) . (P +(Q+R)(Q+\bar{R}))$

$(P + \bar{Q} + \bar{R}) . (P + Q)$

$(P + (\bar{Q}+ \bar{R})Q)$

$P+Q \bar{R}$

Hence,option(B)$P+Q \bar{R}$ is the correct choice.

edited
ans is B

(p+q'+r')(p+q+r)(p+q+r')=(p+q'+r')(p+q)                    (since a=(a+b)(a+b') so putting p+q as a here)

=(p+q')(p+q) +(p+q)r' =p+pr'+qr'=p(1+r')+qr'=p+qr'     (again using above law and 1+anything=1

Another way to solve this problem is as follow -->

+1
0,1 and 3 should be filled with 0 right?
0
yes
0,1 and 3 must be 0.
(P+Q'+R').(P+Q+R).(P+Q+R') is Standard POS which is eqt to ∐ (0,1,3) .

So Standard SOP =∑(2,4,5,6,7)  when we simplified by K-Map it will gives (P+QR').

Hence (B) (P+QR')  is Correct option.
0
thank you ...
+1 vote

## P+QR¯

(p+q'+r')=011=3
(P+q'+r)=010=2
(P+q+r')=001=1
So, π=(1,3,2)
Sop form for minterm is (0,4,5,6,7)

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