Question

If $S_1=\left\{1,2,3,.....,20\right\},S_2=\left\{a,b,c,d\right\},S_3=\{b,d,e,f\};$

The number of elements of $\left(S_1*S_2\hspace{0.17em}\right)\cup \hspace{0.17em}\left(S_1*S_3\right)$ is

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

Answer 1

The correct option is B

$120$

Explanation for the correct option

We know that the number of elements in $S_1$ are $20$ and in $S_2$ there are $4$ elements, then we have

$\begin{array}{rcl}\left(S_1*S_2\hspace{0.17em}\right)& =& 20*4\\ & =& 80\text{elements}\end{array}$

Also, $S_3$ have $4$ elements, then

$\begin{array}{rcl}\left(S_1*S_3\hspace{0.17em}\right)& =& 20*4\\ & =& 80\text{elements}\end{array}$

Now, to find the required, we need to check if there are any common elements.

We can see that the sets $S_2\text{and}{S}_3$ have two common elements which are $b\text{and}d$.

Then, the number of common elements will be,

$20*2=40$

So, the given expression's number of elements are

$\begin{array}{rcl}\left(S_1*S_2\hspace{0.17em}\right)\cup \hspace{0.17em}\left(S_1*S_3\right)& =& \left(S_1*S_2\hspace{0.17em}\right)+\hspace{0.17em}\left(S_1*S_3\right)-\text{number of common elements}\\ & =& 80+80-40\\ & =& 160-40\\ & =& 120\mathrm{elements}\end{array}$

Hence, the correct option is (B).