GO Classes CS Test Series 2025 | Discrete Mathematics | Topic Wise Test 1 | Question: 14

Answer 1

P = ab + bc.

It is given that a, b, c, d are propositional variables.

For P to be true either ab = True or bc = True or both.

For ab = True, the value of a = True (1 choice), b = True (1choice) but value of c and d doesn’t matter. So, c has 2 choices and d also has 2 choices. Therefore, total choices = 1*1*2*2 = 4.

Similarly for bc = True, the value of b = True (1 choice), c = True (1choice) but value of a and d doesn’t matter. So, a has 2 choices and d also has 2 choices. Therefore, total choices = 1*1*2*2 = 4.

So, total = 4 + 4. But note that both the cases can happen together (i.e., both the cases are not disjoint). We have, thus, over-counted. Let’s find the over-count and subtract it from total calculated.

Both cases together True mean a = True, b = True, c = True, d = True/False i.e, total choices = 1*1*1*2 = 2.

Hence, P is true for (8-2) rows = 6 rows. It means there are 6 models for P.

Note that the over-count was counted exactly twice (one time in each case) and so, we subtracted it once from total.

Comments

It is given that $ab+bc$ is true then $b=True$ only because if it is $False$ then whole expression is $False$. Now, after putting $b=True$ we have $a+c=True$ which is true  on $(T,T),(T,F)$ or $(F,T)$. So, total $3$ possibilities and $d$ is not present in the expression. So, it will take any value True or False. So,

=(#of possibilities of b) * (# of possibilities of a and c )*(# of possibilities of d)

=$1*3*2 =6$

Answer 2

Given function $ f(a,b,c,d) = (a \wedge b) \vee (b \wedge c)$

Number of combinations in the 4 variable truth table for which $(a \wedge b)$ is True = 4

Number of combinations in the 4 variable truth table for which $(b \wedge c)$ is True = 4

Number of combinations in the 4 variable truth table for which $((a \wedge b) , (b \wedge c)$ both are True = 2

Apply inclusion exclusion principle,

4 + 4 -2 = 6(ans)