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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).

in Digital Logic by Active (3.8k points) | 60 views

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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 =  $2^{10} - 3^{4}$

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