Question

a,b,c,d are digits in 2014 such that (a,b,c,d}={2,0,1,4}. Then the number of different values ((a^b)^c)^d takes is:-
(a)2
(b)16
(c)8
(d)7

Answer 1

So putting a = 0 we get only 1 value 

now putting b = 0 we get only 1 value 

putting c = 0  a and b can take up values in 3 ways

now putting d = 0 we see we get ³P₂ values out of which we got 3 values already 

so total values ³P₂ - 3 + 2 + 3 = 3! + 2 = 3.2  + 2 = 8 ways