exam approach
m=3 then {1,2,3}
n=1 as there is only one set of size three.
Summation of Sets(containing 1 , containing 2 , containing 3) = Summation of Sets(1,1,1) = 3
A)3*3=9 B)3*1=3{this might be correct answer} C)2*3+1= 7 D)2*1+1 = 3{this might be correct answer}
so lets check for
m=4 then {1,2,3,4}
n=4 as {1,2,3} & {2,3,4} & {1,2,4} & {1,3,4}
Summation of Sets(containing 1 , containing 2 , containing 3,containing 4) = Summation of Sets(3,3,3,3) =12
A)3*4 =12{this might be correct answer} B)3*4={this might be correct answer} C)2*4+1 =9 D) 2*4+1 = 13
Hence B correct