Question

Let $A$ & $B$ be two sets containing $5$ and $2$ elements respectively. Then the number of subsets of $A\times B$ each having atlest $8$ elements is:

• A. $48$
• B. $50$
• C. $52$
• D. $56$

Answer 1

The correct option is D $56$

We have,

$A$ and $B$ two sets containing $5$ and $2$ elements respectively.

Let $A=(1,2,3,4,5)$

$B=(a,b)$

So,

$A\times B=\{(1,a),(1,b),(2,a),(2,b),(3,a),(3,b),(4,a),(4,b),(5,a),(5,b)\}$

Then,$A\times B$ contains $10$ order pairs.

So, we start making the subsets with $10$ pairs.

Now,

With $10$ ordered pairs $=1$ subset

With $9$ ordered pairs${=}^{10}{C}_9=10$ subsets.

With $8$ ordered pairs${=}^{10}{C}_8=45$subsets.

Now,

By adding all the above subsets, we get

$1+10+45=56subsets.$

Hence, this is the answer.