Single Choice

Let X be a set of $$5$$ elements. The number d of ordered pairs (A, B) of subsets of X such that $$A\neq \Phi, B\neq \Phi, A\cap B=\Phi$$ satisfies.

A$$50\leq d \leq 100$$
B$$101\leq d \leq 150$$
C$$151 \leq d \leq 200$$
Correct Answer
D$$201 \leq d$$

Solution

Total no. of ordered pairs is
$${ =C }_{ 2 }^{ 5 }\times 2!+{ C }_{ 3 }^{ 5 }\left( \frac { 3! }{ 1!\times 2! } \times 2! \right) +{ C }_{ 4 }^{ 5 }\left( \frac { 4! }{ 1!\times 3! } \times 2!+\frac { 4! }{ 2!\times 2! } \times \frac { 2! }{ 2! } \right) +{ C }_{ 5 }^{ 5 }\left( \frac { 5! }{ 1!\times 4! } \times 2!+\frac { 5! }{ 2!\times 3! } \times 2! \right) \\ =10(2)+10(6)+5(8+6)(10+20)\\ =180\\ $$
So option $$C$$ is correct


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