Question

If $S_1=\left\{1,2,3,\dots ..20\right\},S_2=\left\{a,b,c,d\right\},\text{and}{S}_3=\left\{b,d,e,f\right\}$. The number of elements of $\left(S_1\times S_2\right)\cup \left(S_1\times S_3\right)$ is

• A. $100$
• B. $120$
• C. $140$
• D. $40$

Answer 1

The correct option is B

$120$

Explanation for the correct option

We will first find the element in $S_1\times S_2$,

As there are twenty element in $S_1$ and four elements in $S_2$, then the elements in $S_1\times S_2$ will be,

$\begin{array}{rcl}\therefore S_1\times S_2& =& 20\times 4\\ & =& 80\end{array}$

Similarly, the elements in $S_1\times S_3$ will be

$\begin{array}{rcl}\therefore S_1\times S_3& =& 20\times 4\\ & =& 80\end{array}$

Now, we need to find the common elements in $S_1\times S_2\text{and}{S}_1\times S_3$ will be the elements with $b\text{and}d$ because they both lie in $S_2\text{and}{S}_3$. Then

$\begin{array}{rcl}\text{Common Elements}& =& 20\times 2\\ & =& 40\end{array}$

Then, the required expression will be,

$\begin{array}{rcl}\therefore \left(S_1\times S_2\right)\cup \left(S_1\times S_3\right)& =& S_1\times S_2+S_1\times S_3-\text{Common Elements}\\ & =& 80+80-40\\ & =& 120\end{array}$

Hence, the correct option is (B).