**order of function P R Q weight of P=4,Q=2,R=1)**

**absent variable we can take 0 and 1 both possibility **

**total possible no **

**PQ=(1110,111)=6,7**

**QR'=(010,110)=2,6**

**PR'=(100,110)=4,6**

**so min term (2,4,6,7)**

**option A**

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+18 votes

The minterm expansion of $f(P,Q,R) = PQ +Q \bar{R}+P\bar{R}$ is

- $m_2+m_4+m_6+m_7$
- $m_0+m_1+m_3+m_5$
- $m_0+m_1+m_6+m_7$
- $m_2+m_3+m_4+m_5$

+30 votes

Best answer

0

but if take the counting of cell of k-map in vertical order the answer is comming; m1+m4+m5+m7

and dont match any options. please explain

and dont match any options. please explain

+2

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)]

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