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Archangel Macsika

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

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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.


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