1 vote

Consider the Boolean function F(x1, x2, . . . , x10) realised by the following combinational circuit.

Determine the number of input combinations for which the output function F realised by the circuit becomes true (logic 1).

1 vote

Best answer

Answer : $2^{10} - 3^{4}$

number of input combinations for which $F$ becomes True = Total Combinations - number of input combinations for which $F$ becomes False

Total Combinations = $2^{10}$

**Number of input combinations for which $F$ becomes False(logic 0) :**

To make $F$ false, all the inputs to the Final OR Gate must be False. Thus, Pair wise Possibilities :

$(x_1,x_2)$ = 1 Possibility (Both must be True to make the Output False)

$(x_3,x_4)$ = 3 Possibility (At least One must be False)

$(x_5,x_6)$ = 3 Possibility (At least One must be True)

$(x_7,x_8)$ = 3 Possibility (At least One must be False)

$(x_9,x_{10})$ = 3 Possibility (At least One must be False)

Thus, Number of input combinations for which $F$ becomes False(logic 0) : $1*3*3*3*3 = 3^4$

So, Final Desired Answer = $2^{10} - 3^{4}$