Let $S=\left\{1,2,3,4\right\}.$ The total number of unordered pairs of disjoint subsets S is equal to

The total number of ordered pairs of disjoint subsets of S , except ordered pair $(\varphi ,\varphi )$ is

$\left({}^{4}C{}_0\times 2^4+{}^{4}C{}_1\times 2^3+{}^{4}C{}_2\times 2^2\times {}^{4}C{}_3\times 2^1+{}^{4}C{}_4\times 2^0\right)–1$

$={(1+2)}^4–1=80$

$\therefore$   Total number of unordered pairs of disjoint subsets of S is equal to $\frac{80}{2}+1$ $=41$ .