Let |X| denote the number of elements in set X . Let S = \ 1, 2, 3,...

Since $A$ and $B$ are independent events
$P\left(A \cap B\right) = P\left(A\right)P\left(B\right)$
$\dfrac{\left|A \cap B\right|}{\left|S\right|} = \dfrac{\left|A\right|}{\left|S\right|} \cdot \dfrac{\left|B\right|}{\left|S\right|}$
$\begin{cases}\left|S\right|\left|A \cap B\right| = \left|A\right| \cdot \left|B\right| & \left(∴ \left|x\right| = numberofelementinset\right)\end{cases}$
\left|S\right| = 6$<br />$6\left|A \cap B\right| = \left|A\right|\left|B\right|$<br />$\begin{cases}6n\left(A \cap B\right) = n\left(A\right)n\left(B\right) & \left(set\left|x\right| = n\left(x\right)\right)\end{cases}$<br />Case$I:$<br />$n\left(A \cap B\right) = 1$<br />$⇒ n\left(A\right)n\left(B\right) = 6$<br />\left(a\right)$\begin{cases}n\left(A\right) = 6, n\left(B\right) = 1 & Numberofways = C6 6 \times C1 6 = 6\end{cases}</p><p>\left(b\right)n\left(A\right) = 3, n\left(B\right) = 1Numberofways = C3 6 \times C1 3 \times C2 5 = 180$</p><p><br />Case $II:$<br />$n\left(A \cap B\right) = 2$<br />$⇒ n\left(A\right)n\left(B\right) = 12$<br />$\begin{cases}\left(a\right)n\left(A\right) = 6 & n\left(B\right) = 2 & Numberofways\end{cases} = {C}_{6} 6 {C}_{2} 6 = 15$<br />$\begin{cases}\left(b\right)n\left(A\right) = 4 & n\left(B\right) = 3 & Numberofways =\end{cases} C4 6 \times C2 4 \times C1 2 = 180$<br />Case $III:$<br />$n\left(A \cap B\right) = 3$<br />$⇒ n\left(A\right)n\left(B\right) = 18$<br />\left(a\right)$\begin{cases}n\left(A\right) = 6, & n\left(B\right) = 3Numberofways ⇒ & C6 6 \times C3 6 = 20\end{cases}
Case $IV:$
$n\left(A \cap B\right) = 4$
$⇒ n\left(A\right)n\left(B\right) = 24$
$\begin{cases}\left(a\right)n\left(A\right) = 6 & n\left(B\right) = 4Numberofways ⇒ & C6 6 \times C4 6 = 15\end{cases}$
Case $V:$
$n\left(A \cap B\right) = 5$
$⇒ 1n\left(A\right)n\left(B\right) = 30$
$\begin{cases}\left(a\right)n\left(A\right) = 6 & n\left(B\right) = 5Numberofways ⇒ & C6 6 C5 6 = 6\end{cases}$
Total ordered pairs $= 422$