Question

Let A and B be two sets such that : n (A) = 20 , $n(A\cup B)=42andn(A\cap B)=4$. Find

(i) n (B) (ii) n (A - B)

(iii) n (B - A)

Answer 1

(i) n (A) = 20, $n(A\cup B)=42andn(A\cap B)=4,$ to find n (B)

We know $n(A\cup B)=n\left(A\right)+n\left(B\right)-(A\cap B)$

$\Rightarrow 42=20+n\left(B\right)-4$

$\Rightarrow 42=16+n\left(B\right)$

$\Rightarrow n\left(B\right)=42-16$

= 26

$\therefore n\left(B\right)=26$

(ii) To find : n (A - B)

We know that if A and B are disjoint sets, then

$A\cap B=\varphi$

$\therefore n(A\cup B)=n\left(A\right)+n\left(B\right)-(A\cap B)$

= $n\left(A\right)+n\left(B\right)-n\left(\varphi \right)$

$\Rightarrow n(A\cup B)=n\left(A\right)+n\left(B\right)$ [$\therefore n\left(\varphi \right)=0$]

Now,

$A=(A-B)\cup (A\cap B)$

i.e. A is the disjoint union A - B and $A\cap B$

$\therefore n\left(A\right)=n(A-B)\cup (A\cap B)$

$=n(A-B)+n(A\cap B)$ [$\because A-BandA\cap B$ are disjoint]

$\Rightarrow 20=n(A-B)+4$

$\Rightarrow n(A-B)=20-4$

= 16

$\therefore n(A-B)=16$

(iii) To find : B - A

On a similar we have B is the disjoint union of B - A and $A\cap B$ i.e. $B=(B-A)\cup (a\cap B)$

$\therefore n\left(B\right)=n(B-A)+n(A\cap B)$

$\Rightarrow 26=n(B-A)+4$ [ using (i)]

$\Rightarrow n(B-A)=26-4$

= 22

$\therefore n(B-A)=22$