Question

Let $\mathrm{A}$ and $\mathrm{B}$ be two sets having $4$ and $8$ elements respectively. Write the minimum number of elements that $\mathrm{A}\cup \mathrm{B}$ can have.

Answer 1

Given, a set $\mathrm{A}$ having $4$ elements i.e. $\mathrm{n}\left(\mathrm{A}\right)=4$

Set $\mathrm{B}$ having $8$ elements i.e. $\mathrm{n}\left(\mathrm{B}\right)=8$

We know that $\mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})..........\left(1\right)$

From the formula, we can say that $\mathrm{A}\cup \mathrm{B}$ having the minimum number of elements when $\mathrm{A}\cap \mathrm{B}$ having the maximum number of elements.

Since $\mathrm{n}\left(\mathrm{A}\right)=4$ and $\mathrm{n}\left(\mathrm{B}\right)=8$, therefore the maximum number of elements in $\mathrm{A}\cap \mathrm{B}$ are $4$.

Substitute all the values in the equation $\left(1\right)$, and we get;

$\mathrm{n}(\mathrm{A}\cup \mathrm{B})=4+8-4=8$

Therefore the minimum number of elements that $\mathrm{A}\cup \mathrm{B}$ can have is $8$.