Question

Let S = {1, 2, 3, 4}.Find the total number of unordered pairs of disjoint subsets of S.

Answer 1

S = {1, 2, 3, 4}. Let us assume S has m elements
No. of ways to select n elements out of $m_1{=}^m{C}_n$
${}^m{C}_n=\frac{m!}{(m-n)!n!}$
Remaining elements = (m - n) [u, z, n are already selected]
No. of subsets that can be formed from (m - n) elements $=2^{m-n}$
So, for any particular value of n, total number of disjoint sets
$=\frac{m!}{(m-n)!n!}\times 2^{m-n}$
because $2^{m-n}$ subsets have all different elements form $=\frac{m!}{(m-n)!n!}$
So, total no. of disjoint ordered sets pair =
${\sum }_{n=0}^4\frac{m!}{(m-n)!n!}\times 2^{m-n}$
$={\sum }_{n=0}^4{}^m{C}_n{2}^{m-n}.1^n$
Comparing with form $(1+x{)}^m={\sum }_{n=0}^m{}^m{C}_n{x}^{m-n}{1}^n$ we get
${\sum }_{n=0}^4{}^m{C}_n{2}^{m-n}{1}^n=(2+1{)}^m=3^m$
There, $3^m$ pairs also contain $(\Phi ,\varphi )$. So, subtracting that leaves $(3^m-1)$ which are ordered. This means for eg. {1}, {2,3,4} and {(2,3,4),(1)} will both be corrected. But in question only unordered pairs are asked. So, total no. of pairs of disjoint unordered sets $=\frac{3^m-1}{2}+\frac{1}{m}$ for $(\Phi ,\varphi )$
$[sina,m=A]=\frac{3^4-1}{2}+1=41pairs$