Answer to Question #142104 in Discrete Mathematics for Promise Omiponle

Question

Let S be a finite set, with |S|= 200. Find the number of subsets of S containing more than 2 elements

Answer 1

The number of subsets of $S$ containing k elements is equal to $\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!(n-k)!}$, where $n=|S|=200.$

Firstly, find the number $m$ of subsets of $S$ containing not more than 2 elements. The emptyset is a unique set containg no elements. There are 200 different subsets of $S$ having one element. The number of subsets containg two elements is $\left(\begin{array}{c}200\\2\end{array}\right)=\frac{200!}{2!\cdot198!}=\frac{200\cdot199\cdot198!}{2\cdot 198!}=100\cdot199=19900.$

Then $m=1+200+19900=20101.$ Since the set $S$ contains $2^{200}$ subsets, the number of subsets of $S$ containing more than 2 elements is $2^{200}-m=2^{200}-20101.$