The output of the AND gate is high only if all the inputs connected to it are high.
So, For the device to get enable, O/p's of all the four gates should be high, So that all the inputs to AND will become high.
O/p of X-NOR will be high only if it gets {00 or 11} as i/p. Hence 2 choices for A0, A1 which are 00 or 11.
For X-or o/p will be high only for {01 or 10} so 2 possible combinations.
Similarly, for NOR, 1 valid combination {00} and for OR 3 valid combinations {01,10,11}
So, Total 2*2*1*3=12 addresses we can form for which device will be enabled.
Option C.
