in Digital Logic retagged by
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in Digital Logic retagged by
321 views

4 Comments

@Manoj, I got 10 posiible cases can u explain how u got 24?
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Vignesh: Y is one when $(A_2 \oplus B_2 \oplus A_1 \oplus B_1 \oplus A_0 \oplus B_0)$ is 0.

so when no of 1's is even. 

That is $^6C_0+ ^6C_2 + ^6C_4 + ^6C_6 = 1+15+15+1= 32$

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Thank You
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1 Answer

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I think here are 30 combination

edited by

3 Comments

Got it mate so the answer is 16 right. Thanks a lot

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@Manojk

in ur 1st combination 3 outputs are coming 0 0 0

0 XNOR 0 XNOR 0 = 0

not 1 we are getting in Y

Similarly 2nd , 3rd , 4th combination too

Am I wrong?
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thanks smiley

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