Digital Logic

Question

A digital circuit shown below has two $3$-bit input $A_2A_1A_0$ and $B_2B_1B_0$.

To obtain output $Y=1$, the number of possible cases are ______.

Comments

https://gateoverflow.in/44664/given-circuit-anyone-tell-possible-input-cases-which-output

Answer 1

Output of XOR, $Y$, $(A_2 \oplus B_2 \oplus A_1 \oplus B_1 \oplus A_0 \oplus B_0)$,  is $1$,  only when we have ODD number of $1's$ in input. 

Possibilities

1. One $1$ out of $6$ inputs $= {}^6C_1 = 6$

2. Three $1's$ out of $6$ inputs $= {}^6C_3 = 20$

3. Five $1's$ out of $6$ inputs $= {}^6C_5 = 6 $

Total possibilities to get $ Y = 1 $ are $6+20+6=32$

Comments

This seems to be better approach, thanks a lot. :)

Answer 2

TO GET y=1 .possible cases are:

001,,,,,,010,,,,100....111

001-> A2 B2=(00,11) ,A1 B1=(00,11) ,A0B0=(10,01).....means we have 2x2x2 = 8 ways ...similarly for all ..

TOTAL 4x8=32

Comments

I could not understand 00,11 what it means? Can you elaborate it a bit please?