Question

Let X = {1, 2, 3, 4, 5, 6}. The number of ways of different ordered pairs (A, B) that can be formed such that A and B are subsets of X and $A\cap B=\phi$, is

• A. $6^2$
• B. $2^6$
• C. $3^6$
• D. $6^3$

Answer 1

The correct option is C $3^6$

Since A, B are subsets of X, Let a $\in X$. There are four possibilities
$1.a\in A,a\in B\Rightarrow a\in A\cap B2.a\in A,a\text{/}\in B\Rightarrow a\text{/}\in A\cap B3.a\text{/}\in A,a\in B\Rightarrow a\text{/}\in A\cap B4.a\text{/}\in A,a\text{/}\in B\Rightarrow a\text{/}\in A\cap B$
In these four cases, only three cases are possible such that $A\cap B=\phi$
$\therefore$ For each element in X has three possible cases.
$\therefore$ The total number of ordered pairs (A, B) such that $A\cap B=\phi$ is equal to $3\times 3\times 3\times 3\times 3\times 3=3^6$