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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

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(a)2

(b)16

(c)8

(d)7

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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