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given the Boolean expression for F(A,B,C)
F(AB,C) = AB+ ABC
. How many binary input comibinations are there for the output funtion F(A,B,C) to be 1


Sagot :

No of variables in Boolean expression is 3 i.e A,B,C

  • n=3

Number of input combinations.

[tex]\\ \sf\longmapsto 2^n[/tex]

[tex]\\ \sf\longmapsto 2^3[/tex]

[tex]\\ \sf\longmapsto 8[/tex]

It will follow the octal table.

000

001

010

011

100

101

110

110

We know the rule .

  • If there is 1 present then output comes 1.
  • Only 000 has no 1s

So

no of outputs which are 1=8-1=7.