Answer : 32

Solution:

XOR gates are used. The output of XOR gate is 1, when inputs have odd number of ones.

i.e For Y=1, we have 4 possible cases: 001,010,100,111

Now for each case: For example consider case 1(001):

->A0⊕B0=1 & A1⊕B1=0 & A2⊕B2=1

-> For A0⊕B0=1 : two possiblities - A0,B0= 0,1 or 1,0

A1⊕B1=0 : two possiblities - A1,B1= 0,1 or 1,0

A2⊕B2=1 : two possiblities - A2,B2= 0,1 or 1,0

Hence total no. of combinations=2*2*2 = 8

Hence for each case: 8 combinations.

Hence total no. of possible cases= 4 * 8 = 32.