Probo from set theory

No one wow!!!!!!

no of elements in m = 6 and n = 3

It's clear that the first set has 2^m subsets and second one has 2ⁿ subests.

so we have to solve 2^m - 2ⁿ = 56 (m,n are positive integers)

Its also clear that m>n

let m = k+n

so 2^k+n- 2ⁿ= 56

2ⁿ(2^k- 1)= 2³*7

clearly 2^k-1 is odd

so 2ⁿ= 2³, and n = 3

and 2^k-1= 7 so k =3

so m = 3+3 = 6 and n = 3

;O)

hit and trial se ho gaya na!!

bas theek hai!!
:P

but good job shubhodip....:)

thnk u :P