Let A has m elements
Let B has n elements
Total number of students of A = \(2^m\)
Total number of students of B = \(2^n\)
It is given: \[ 2^m - 2^n = 56 \] \[ 2^n (2^{m-n} - 1) = 56 \] \[ 2^n = \text{even and } 2^{m-n} - 1 = \text{odd} \] Now, \[ 56 = 8 \times 7 = 2^3 \times 2^7 \] \[ 2^n (2^{m-n} - 1) = 2^3 \times 7 \] \[ \Rightarrow n = 3 \] Now, \[ 8 (2^{m-3} - 1) = 8 \times 7 \] \[ \Rightarrow 2^{m-3} - 1 = 7 \] \[ \Rightarrow 2^{m-3} = 8 = 2^3 \] \[ \Rightarrow m - 3 = 3 \] \[ \Rightarrow m = 6 \]