Question

Let Z be the set of integers. If A = $\left\{ x\in Z:{{2}^{\left( x+2 \right)\left( {{x}^{2}}-5x+6 \right)}}=1 \right\}$ and B = $\left\{ x\in Z:-3<2x-1<9 \right\}$, then the number of subsets of the set A $\times $ B, is:
(a) ${{2}^{18}}$
(b) ${{2}^{10}}$
(c) ${{2}^{15}}$
(d) ${{2}^{12}}$

Answer 1

Hint: To solve this question, we first need to find the sets A and B. We are given the conditions for both the sets. We will find all the possible values of x for both the sets. Once we have A and B, we will find the cartesian product of the two sets represented by A $\times $ B. Then, we will find the number of subsets that can be formed from the elements in the set A $\times $ B.

Complete step-by-step solution:
It is given that Z is the set of all the integers. It is also given that A = $\left\{ x\in Z:{{2}^{\left( x+2 \right)\left( {{x}^{2}}-5x+6 \right)}}=1 \right\}$ and B = $\left\{ x\in Z:-3<2x-1<9 \right\}$.
First of all, we shall find the set A. It is given that x belongs to the set of integers such that ${{2}^{\left( x+2 \right)\left( {{x}^{2}}-5x+6 \right)}}=1$. We know that any number raised to power 0 is 1.
$\Rightarrow \left( x+2 \right)\left( {{x}^{2}}-5x+6 \right)=0$
Therefore, we can say that $x+2=0$ or ${{x}^{2}}-5x+6=0$.
$\Rightarrow x=-2$
The second equation is a quadratic equation. We will solve this equation by factorization method.
In this method, we will split the middle term such that the product of the two-term after splitting is equal to the product of the first term and the constant term.
Hence, we will divide $─5x$ as $─3x$ and $─2x$.
$\begin{align}
  & \Rightarrow {{x}^{2}}-3x-2x+6=0 \\
 & \Rightarrow x\left( x-3 \right)-2\left( x-3 \right)=0 \\
 & \Rightarrow \left( x-3 \right)\left( x-2 \right)=0 \\
\end{align}$
Thus, $x = 3$ or $x = 2.$
Therefore, $x = ─2, 2, 3$
Thus, $A = {─2, 2, 3}$
Now, we will find the set B.
It is given that x belongs to a set of integers such that $-3< 2x-1< 9$.
We will solve this inequality to get all the values of x.
 $\begin{align}
  & \Rightarrow -3+1< 2x-1+1< 9+1 \\
 & \Rightarrow \dfrac{-2}{2}< \dfrac{2x}{2}< \dfrac{10}{2} \\
 & \Rightarrow -1< x< 5 \\
\end{align}$
Therefore, x can be $0, 1, 2, 3, 4$.
Hence, $B = {0, 1, 2, 3, 4}$
Now, we will find the cartesian product of A and B.
Thus A $\times B = {(─2, 0), (─2, 1), (─2, 2), (─2, 3), (─2, 4), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4) , (3, 0), (3, 1), (3, 2), (3, 3), (3, 4)}$
Number of elements in $A \times B$ = $n(A \times B)$ = 15.
Now, we will find the number of subsets of $A \times B$.
The subset can be a null set.
Thus, the number of ways of selecting no elements from a set of 15 elements is given as $^{15}{{C}_{0}}$.
Each subset can have 1 element each.
Thus, the number of ways of choosing 1 element from a set of 15 elements is given as $^{15}{{C}_{1}}$.
Similarly, if each subset has 2 elements, the number of subsets will be $^{15}{{C}_{2}}$.
Therefore, total number of subsets will be $^{15}{{C}_{0}}{{+}^{15}}{{C}_{1}}{{+}^{15}}{{C}_{2}}+....{{+}^{15}}{{C}_{15}}$.
But we know that $\sum\limits_{i}^{n}{^{n}{{C}_{i}}}={{2}^{n}}$
Therefore, total number of subsets will be $^{15}{{C}_{0}}{{+}^{15}}{{C}_{1}}{{+}^{15}}{{C}_{2}}+....{{+}^{15}}{{C}_{15}}={{2}^{15}}$.
Hence, option (c) is the correct option.

Note: To prove that $\sum\limits_{i}^{n}{^{n}{{C}_{i}}}={{2}^{n}}$, consider the binomial expansion of ${{\left( 1+x \right)}^{n}}$. We know that ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( 1 \right)}^{0}}{{x}^{n-0}}{{+}^{n}}{{C}_{1}}{{\left( 1 \right)}^{1}}{{x}^{n-1}}+...{{+}^{n}}{{C}_{n}}{{\left( 1 \right)}^{n}}{{x}^{n-n}}$. Now put x = 1. ${{\left( 1+1 \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( 1 \right)}^{0}}{{\left( 1 \right)}^{n-0}}{{+}^{n}}{{C}_{1}}{{\left( 1 \right)}^{1}}{{\left( 1 \right)}^{n-1}}+...{{+}^{n}}{{C}_{n}}{{\left( 1 \right)}^{n}}{{\left( 1 \right)}^{n-n}}$. Therefore, we can say that ${{\left( 2 \right)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}+...{{+}^{n}}{{C}_{n}}=\sum\limits_{i}^{n}{^{n}{{C}_{i}}}$ .