Question

If $A$ and $B$ be two sets such that $n\left(A\right)=15$, $n\left(B\right)=25$, then the number of possible values of $n(A△B)$ (symmetric difference of $A$ and $B$) is

• A. $30$
• B. $16$
• C. $26$
• D. $40$

Answer 1

The correct option is B

$16$

Explanation for the correct option:

Step 1: Frame a relation between $n(A△B)$ and $n(A\cap B)$

Given, $n\left(A\right)=15$ and $n\left(B\right)=25$.

Symmetric difference of two sets is defined as the difference of the union and intersection of the two sets. i.e.,

$\mathit{n}\mathbf{(}\mathit{A}\mathbf{△}\mathit{B}\mathbf{)}\mathbf{=}\mathit{n}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}\mathit{–}\mathit{n}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}$

It can be rewritten as,

$$\begin{gathered} n(A△B)=n\left(A\right)+n\left(B\right)-n(A\cap B)-n(A\cap B)[\because n(A\cup B)=n(A)+n(B)-n(A\cap B\left)\right] \\ \Rightarrow n(A△B)=n\left(A\right)+n\left(B\right)-2·n(A\cap B) \\ \Rightarrow n(A△B)=15+25-2·n(A\cap B) \\ \Rightarrow n(A△B)=40-2·n(A\cap B)\left(1\right) \end{gathered}$$

Step 2: Calculate range of values of $n(A△B)$

For $n(A△B)$ to be minimum, value of $n(A\cap B)$ should be the maximum.

The maximum possible value of $n(A\cap B)$ is the number of elements in the smaller set.

Here, the smaller set is $B$ since it has only $15$ elements. Thus,

$\mathrm{n}{(A△B)}_{\min }=40-2\times 15=10$.

For $n(A△B)$ to be maximum, value of $n(A\cap B)$ should be the minimum.

The minimum possible value of $n(A\cap B)$ is $0$.

So,

$\mathrm{n}{(A△B)}_{\max }=40-2\times 0=40$

Thus, $n(A\cap B)$ can vary from $0$ to $15$, accordingly , $n(A△B)$ varies from $10$ to $40$ with a difference of $2$ as second multiple of $n(A\cap B)$ is being subtracted.

Thus, the possible values for $n(A△B)=\{10,12,14,16,\dots ,40\}$, obtained by substituting possible values of $n(A\cap B)$ into $\left(1\right)$

Step 3: Calculate the number of elements in the range of values of $n(A△B)$

We can observe that, the elements in the set which holds the possible values of $n(A△B)$ are in an arithmetic progression.

The $n^{\mathrm{th}}$ term of an arithmetic progression is given as,

$a_n=a+(n-1)d$

Where, $a_n$ is the $n^{\mathrm{th}}$ term, $a$ is the first term and $d$ is the difference between each consecutive terms in the series, i.e., common difference.

Here, $a_n=40$, $a=10$ and $d=2$.

Substituting the values in the formula,

$$\begin{gathered} \Rightarrow \\ \Rightarrow \\ \Rightarrow \end{gathered}$$ $\begin{array}{rcl}40& =& 10+(n-1)\times 2\\ n-1& =& 15\\ n& =& 16\end{array}$

Thus, the number of possible values of $n(A△B)=16$

Hence, option B is correct.