Question

Let X = {1, 2, 3, 4, 5}. What is the number of different ordered pairs (Y, Z) that can be formed such that $Y\subseteq X,Z\subseteq X\text{ and }Y\cap Z$ is empty?
(a) ${{5}^{2}}$
(b) ${{3}^{5}}$
(c) ${{2}^{5}}$
(d) ${{5}^{3}}$

Answer 1

Hint: We are going to divide the problem into 6 cases and the case will be on the basis of the number of elements in Y and by using this we have the number of elements in Z , with which we will find the ordered pair and add it for each of the cases.

Complete step-by-step answer:

Let’s start by taking cases:
We are going to use these formula:
For a set of size k the number of subsets is ${{2}^{k}}$ .
Case 1: Set Y has 0 elements, then there are ${}^{5}{{c}_{0}}$ choices of Y here.
So, Z has 5 elements and for each Y there are ${{2}^{5}}$ possibilities for Z.
Therefore, in this case we have ${}^{5}{{c}_{0}}$.${{2}^{5}}$ possibilities.
Case 2: Set Y has 1 elements, and similarly we have
${}^{5}{{c}_{1}}{{.2}^{4}}$ possibilities,
Case 3: Set Y has 2 elements, and similarly we have
${}^{5}{{c}_{2}}{{.2}^{3}}$ possibilities,
Case 4: Set Y has 3 elements, and similarly we have
${}^{5}{{c}_{3}}{{.2}^{2}}$ possibilities,
Case 5: Set Y has 4 elements, and similarly we have
${}^{5}{{c}_{4}}{{.2}^{1}}$ possibilities,
Case 6: Set Y has 5 elements, and similarly we have
${}^{5}{{c}_{5}}{{.2}^{0}}$ possibilities,
Now the total possibility is,
$={}^{5}{{c}_{0}}{{.2}^{5}}+{}^{5}{{c}_{1}}{{.2}^{4}}+{}^{5}{{c}_{2}}{{.2}^{3}}+{}^{5}{{c}_{3}}{{.2}^{2}}+{}^{5}{{c}_{4}}{{.2}^{1}}+{}^{5}{{c}_{5}}{{.2}^{0}}$
Now we will use the formula,
${{\left( 1+x \right)}^{n}}={}^{n}{{c}_{0}}{{x}^{n}}+{}^{n}{{c}_{1}}{{x}^{n-1}}+......+{}^{n}{{c}_{n}}{{x}^{0}}$
Using this formula we get,
$\begin{aligned} & {}^{5}{{c}_{0}}{{.2}^{5}}+{}^{5}{{c}_{1}}{{.2}^{4}}+{}^{5}{{c}_{2}}{{.2}^{3}}+{}^{5}{{c}_{3}}{{.2}^{2}}+{}^{5}{{c}_{4}}{{.2}^{1}}+{}^{5}{{c}_{5}}{{.2}^{0}} \\\ & ={{\left( 1+2 \right)}^{5}} \\\ & ={{3}^{5}} \\\ \end{aligned}$
Hence, option (b) is correct.

Note: We can also solve this question by considering that for each member of X we have three choices for it: put it in set Y, put it in set Z, or leave it out. That also gives us the same answer.