# Solve and explain the answer

Dear student
 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
${=}^{5}{C}_{0}×{2}^{5}{+}^{5}{C}_{1}×{2}^{5-1}+\dots {+}^{5}{C}_{5}×{2}^{5-5}$
$=\left(2+1{\right)}^{5}={3}^{5}$
Regards

• 0
What are you looking for?