Question

Let $A$and $B$ be two sets containing $2$ elements and $4$ elements, respectively. The number of subsets of $AxB$ having $3$ or more elements, is equal to?

• A. $256$
• B. $220$
• C. $219$
• D. $211$

Answer 1

The correct option is C

$219$

Determine the number of subsets of $AxB$ having $3$ or more elements

Total number of elements in $AxB$ is the multiplication of the number of elements in $A$ and number of elements in $B$

Number of elements in $AxB$$=2\times 4=8$.

Number of a subset in $AxB$ $=2^n=2^8$

Number of subsets with zero element is $8{\mathrm{ℂ}}_0$

Number of subsets with one element is $8{\mathrm{ℂ}}_1$

Number of subsets with two elements is $8{\mathrm{ℂ}}_2$

Number of subsets of $AxB$ having $3$ or more element is equal to

$$\begin{gathered} \Rightarrow 2^8-8{\mathrm{ℂ}}_0-8{\mathrm{ℂ}}_1-8{\mathrm{ℂ}}_2 \\ \Rightarrow 2^8-\left[\frac{8!}{\left(8-0\right)!0!}+\frac{8!}{\left(8-1\right)!1!}+\frac{8!}{\left(8-2\right)!2!}\right]\left[\because n{\mathrm{ℂ}}_r=\frac{n!}{\left(n-r\right)!r!}\right] \\ =2^8-\left[\frac{8!}{\left(8\right)!}+\frac{7!\times 8}{\left(7\right)!}+\frac{8!}{\left(6\right)!2!}\right]\left[\because 0!=1,1!=1\right] \\ =2^8-\left[1+8+\frac{7\times 8}{2}\right] \\ =256-\left[1+8+28\right] \\ =219 \end{gathered}$$
Hence, option (C) is the correct answer.