Let N be the number of ordered pairs of non-empty sets A and B that have the following properties

$\text{ i) }A\cup B=\{1,2,3,4,5,6,7,8,9,10,11,12\}$

$\text{ ii) }A\cap B=\varphi$

iii) The number of elements of A is not an element of A

iv) The number of elements of B is not an element of B. Then N is equal to

$\text{ clearly }A\text{ and }B\text{ are two non empty sets }$

$\text{ Case (i) if A contains }1\text{ element then B contains }11\text{ elements }$

$\begin{array}{l}\Rightarrow 1\notin A\text{ and }11\notin B\\ \Rightarrow A=\left\{11\right\},B=\{1,2,3,4\dots ,10,12\}\\ \text{ i.e., only one way }\end{array}$

only one way

       Case (ii) if A contains 2 element then B contains 10 elements

$\Rightarrow 10\in A\text{ and }2\in B$

$\text{ The second element in }\mathrm{A}{\text{ we can select in }}^{10}{C}_1\text{ ways }$

$\underset{\_}{\text{ Similarly if }}$

$n\left(A\right)=3{\text{ then number of possibilities is }}^{10}{C}_2$

$n\left(A\right)=4{\text{ then number of possibilities is }}^{10}{C}_3$

$n\left(A\right)=5{\text{ then number of possibilities is }}^{10}{C}_4$

$n\left(A\right)=6\text{ then }n\left(B\right)=6$

$\text{ Hence }6\notin A,6\notin B\text{ Which is FALSE }$

$\therefore \text{ required number of order pairs is }\left({ }^{10}{C}_0{+}^{10}{C}_1{+}^{10}{C}_2+\dots .{.}^{10}{C}_{10}\right){-}^{10}{C}_5$

$=2^{10}-{}^{10}{C}_5=772$