Question

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A×B having 3 or more elements is:
$
  {\text{A}}{\text{. 200}} \\
  {\text{B}}{\text{. 219}} \\
  {\text{C}}{\text{. 211}} \\
  {\text{D}}{\text{. 256}} \\
$

Answer 1

Hint: In order to find the number of subsets A × B has, we find out how many possible subsets the number of elements in A×B can form. From the total number of subsets we subtract the number of subsets formed by zero, one and two elements to determine the answer.

Complete step-by-step answer:
Given Data,
A has 2 elements
B has 4 elements

According to the given condition, let us consider the sets A and B containing 2 and 4 elements respectively as follows:
Let A = {x, y}
B = {p, q, r, s}

By A×B the question means forming a new set of ordered pairs which has all the elements containing in both A and B respectively.
The number of elements in it is given by the product of individual elements in A and B.
Total number of elements in A×B = 2 × 4 = 8.

The number of subsets formed by a given set A×B is given by${{\text{2}}^{\text{n}}}$, where n is the number of elements in the set.
The number of subsets in A×B = ${{\text{2}}^8}$= 256

Now the number of ways we get favorable outcomes from a set of ‘n’ total outcomes is found by calculating combination which is given by the formula,
${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!{\text{r!}}}}$

Now the number of subsets having 3 or more elements = Total number of possible subsets – (number of subsets having zero elements + one element + two elements)
Number of subsets having zero elements is given by = ${}^8{{\text{C}}_{\text{0}}} = \dfrac{{{\text{8!}}}}{{\left( {{\text{8 - 0}}} \right)!0{\text{!}}}} = 1$
Number of subsets having one elements is given by = ${}^8{{\text{C}}_1} = \dfrac{{{\text{8!}}}}{{\left( {{\text{8 - 1}}} \right)!1{\text{!}}}} = 8$
Number of subsets having two elements is given by = ${}^8{{\text{C}}_2} = \dfrac{{{\text{8!}}}}{{\left( {{\text{8 - 2}}} \right)!2{\text{!}}}} = 28$

Therefore the number of subsets having 3 or more elements = 256 – 1 – 8 – 28 = 219.
Option B is the correct answer.

Note: In order to solve this type of problems the key is to know the concept of subsets and the definition of A×B for two sets. Knowing these concepts help us determine the total number of elements we are dealing with, then we can start counting the possible outcomes. The formula of combination and the way of computing the possible outcomes should be done carefully. The factorial of a number n, represented as n! is given by n! = n (n-1)(n-2) …… (n - (n-1)).