Solve and explain the answer

Dear student
$$\begin{gathered} \mathrm{A} \mathrm{set} \mathrm{X}=\left\{1,2,3,4,5\right\} \\ \mathrm{To} \mathrm{ifnd}: \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{different} \mathrm{ordered} \mathrm{pairs} (\mathrm{Y},\mathrm{Z}) \mathrm{such} \mathrm{that} \mathrm{Y}\subseteq \mathrm{X},\mathrm{Z}\subseteq \mathrm{X} \mathrm{and} \mathrm{Y}\cap \mathrm{Z}=\mathrm{\varphi }. \\ \mathrm{Since}, \mathrm{Y}\subseteq \mathrm{X},\mathrm{Z}\subseteq \mathrm{X}, \mathrm{hence} \mathrm{we} \mathrm{can} \mathrm{only} \mathrm{use} \mathrm{the} \mathrm{elements} \mathrm{of} \mathrm{X} \mathrm{to} \mathrm{construct} \mathrm{sets} \mathrm{Y} \mathrm{and} \mathrm{Z}. \end{gathered}$$

n(Y)	Number of ways to make Y	Number of ways to make Z such that $Y\cap Z=\varphi$
0	${}^{5}C_0$	32
1	${}^{5}C_1$	16
2	${}^{5}C_2$	8
3	${}^{5}C_3$	4
4	${}^{5}C_4$	2
5	${}^{5}C_5$	0

Let us explain anyone of the above 6 rows ray third row. In third row,
Number of elements in Y=2
Number of ways to select Y=${}^{5}C_2$ ways.
Because any 2 elements of X can be part of Y.
Now, if Y contains any 2 elements, then these 2 elements cannot be used in any way to construct Z because we want Y U Z = ϕ. And from the remaining 3 elements which are not present in Y, 2³ subsets can be made each of which can be equal lo Z and still Y ⋂ Z = ϕ will be true.  Hence, total number of ways to construct sets Y and Z such that Y ⋂ Z = ϕ
${=}^5{C}_0\times 2^5{+}^5{C}_1\times 2^{5-1}+\dots {+}^5{C}_5\times 2^{5-5}$
$=(2+1{)}^5=3^5$
Regards