<s>[INST] ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is 
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is

$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
then total no. of Subest of $X = 2^5 = 32$
(Which also contain $\phi$)
$\bullet\; $ If $A = \{1\}\;,$ Then $B=\left (\{2\}\;,\{3\}\;,\{4\}\;,\{5\}\;,\{2,3\}\;,\{2,4\}\;,\{2,5\}\;,\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{2,3,4\}\;,\{2,3,5\}\;,\{3,4,5\}\;,\{2,4,5\}\;,\{2,3,4,5\}\right )$
Similarly for $A=\{2\}\;,A=\{3\}\;,A=\{4\}$ and $A=\{5\}\;,$ we get $15$ ordered pair for each single
element ed set $A$
$\bullet\; $ If $A = \{1,2\}\;,$ Then $B=\left(\{3\}\;,\{4\}\;,\{5\},\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{3,4,5\}\right)$
Similarly for $A=\{1,3\}\;,A=\{1,4\}\;,A=\{1,5\}\;,A=\{2,3\}\;,A=\{2,4\}\;,A=\{2,5\}\;,A=\{3,4\}\;,A=\{3,5\}\;,A=\{4,5\}$ we get $7$ ordered pair for each double elemented set $A$
$\bullet\; $ If $A=\{1,2,3\}\;,$ Then $B=\left(\{4\}\;,\{5\}\;,\{4,5\}\right)$
So for $10$ ordered pair of $A\;,$ we get $3$ ordered pair of $B$
$\bullet \;$ If $A = \{1,2,3,4\}\;,$ Then $B = \{5\}$
So for $5$ ordered pair of $A\;,$ we get $1$ ordered pair of $B$
So Total ordered pair of $\left(A,B\right)$ is $ = \left(5\times 15\right)+\left(10 \times 7\right)+\left(10 \times 3\right)+\left(5 \times 1\right) = 75+70+30+5 = 180$
If Question is How can we solve using Combination (selection) way.
plz explain me, Thanks
 [/INST] First let us try to get the number of ordered pairs $(A,B)$ such that $A\cup B = \phi$, without any restrictions on $A$ and $B$. Clearly, each element can EITHER be in only $A$ OR only $B$ OR neither. So, that is $3^5$. Now, remove the cases when $A = \phi$. This is $2^5-1$. Similarly for $B$. So, the total is $3^5 - 2^5 - 2^5 + 1  = 180$ (the +1 is because we have counted the pair $(A,B) = (\phi,\phi)$ twice).
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