Question

Let A and B be two sets containing 4 and 2 elements respectively. Then the number of subsets of the set $A\times B$ each having at least three elements is
A. 219
B. 235
C. 228
D. 256

Answer 1

To solve this question, we will first calculate the number of elements in Cartesian product of $A\times B$ which is calculated as
$n\left( A\times B \right)=n\left( A \right)\times n\left( B \right)$
Where n (A) represents number of elements in set A. Then, we will calculate number of subset of $A\times B$ using formula ${{2}^{n\left( A\times B \right)}}$
Finally, we will subtract the number of subsets of $A\times B$ having 0, 1 and 2 elements from the total number of subsets of $A\times B$ to get our result.

Complete step-by-step answer:
We are given two sets A and B.
If n (A) represents number of elements in set A then,
Given $n\left( A \right)=4\text{ and }n\left( B \right)=2$
As the number of elements in A was 4 and number of elements in B was 2.
Before solving further let us first define $A\times B$
$P\times Q$ is the Cartesian product of two sets P and Q which is given as
P\times Q=\left\\{ \left( p,q \right):p\in P\text{ and }q\in Q \right\\}
Number of elements in $P\times Q\Rightarrow n\left( P\times Q \right)$ is calculated by $n\left( P\times Q \right)=n\left( P \right)\times n\left( Q \right)$
That is, by product of number of elements in P and number of elements in Q.
And the formula to calculate total number of subsets of $P\times Q$ is given by ${{2}^{n\left( P\times Q \right)}}$
Here, we have $n\left( A \right)=4\text{ and }n\left( B \right)=2$
Then, $n\left( A\times B \right)$ can be calculated using the above stated formula.
Doing so, we get:

& n\left( A\times B \right)=n\left( A \right)\times n\left( B \right) \\\ & n\left( A\times B \right)=4\times 2 \\\ & \Rightarrow n\left( A\times B \right)=8 \\\ \end{aligned}$$ Again the total number of subsets of $A\times B$ can be calculated using above formula; $$\begin{aligned} & \text{Total number of subsets of }A\times B={{2}^{n\left( A\times B \right)}} \\\ & \Rightarrow {{2}^{8}} \\\ \end{aligned}$$ So, we have total number of subsets of $$\Rightarrow A\times B={{2}^{8}}$$ Now, we will calculate the number of subsets having at least 3 elements. This can be obtained by subtracting the case when subsets of $A\times B$ has 0, 1 and 2 elements from total number of subsets of $A\times B$ Consider the Case I: when the number of subsets of $A\times B$ has 0 elements. Then, number of subsets of a set having n elements where r is number of elements of subset is given by ${}^{n}{{C}_{r}}$ Therefore, number of subsets of $A\times B$ having 0 elements is given by ${}^{8}{{C}_{0}}$ Consider Case II: when a subset of $A\times B$ has 1 element. Then, number of subsets of $A\times B$ having 1 element is ${}^{8}{{C}_{1}}$ Similarly, number of subsets of $A\times B$ having 2 elements is ${}^{8}{{C}_{2}}$ Therefore, the number of subsets having at least 3 elements is $$\Rightarrow {{2}^{8}}-{}^{8}{{C}_{0}}-{}^{8}{{C}_{1}}-{}^{8}{{C}_{2}}$$ Using the formula: $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ Number of subsets having 3 elements is; $$\begin{aligned} & \Rightarrow {{2}^{8}}-1!-\dfrac{8\times 7!}{7!}-\dfrac{8\times 7\times 6!}{6!\times 2!} \\\ & \Rightarrow {{2}^{8}}-1-8-28 \\\ & \Rightarrow {{2}^{8}}-37 \\\ \end{aligned}$$ Expanding ${{2}^{8}}$ we get: $$\begin{aligned} & \Rightarrow 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2-37 \\\ & \Rightarrow 256-37 \\\ & \Rightarrow 219 \\\ \end{aligned}$$ Therefore, the number of subsets of $A\times B$ having at least 3 elements is 219, **So, the correct answer is “Option A”.** **Note:** The biggest possibility of mistake in this question can be while calculating number of subsets of $A\times B$ having 0 elements, it will be ${}^{8}{{C}_{0}}$ and not ${}^{{{2}^{8}}}{{C}_{0}}$ because 8 is total number of elements in $A\times B$ and ${{2}^{8}}$ is total subset of $A\times B$. Observe clearly that while calculating a subset of $A\times B$ having 0 elements. We have two elements from the set $A\times B$ which has 8 elements and not ${{2}^{8}}$.