Correct option (D) 22n
The number of subsets of the set which contain at most n elements is
2n+1C0 + 2n+1C1 + 2n+1C2 + … + 2n+1Cn = N (say)
We have
2N = 2(2n+1C0 + 2n+1C1 + 2n+1C2 + …+ 2n+1Cn)
= (2n+1C0 + 2n+1C2n+1) + (2n+1C1 + 2n+1C2n) + … + (2n+1Cn + 2n+1Cn+1)
(Since, nCr = nCn–r)
= 2n+1C0 + 2n+1C1 + 2n+1C2 + … + 2n+1C2n+1 = 22n+1
⇒ N = 22n