Explanation:

Explanation:

B Given,
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=99\)
Find, \(\quad \mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\) ?
Then,
\(\mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\mathrm{n}\{(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})\}\)
\(=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=99 \times 99=99^2\)

Explanation:

A Number of element in set \(\mathrm{A}=5\)
And no. of element in set \(B=2\)
The no. of element in ordered pair \(A \times B=2 \times 5=10\)
\(\mathrm{n}(\mathrm{A}+\mathrm{B})=10\)
Then, The number of subsets of \(A \times B\) each having at least 3 and at most 6 elements is-
\(={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6\)
We know that,
\({ }^n c_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{10 !}{3 ! \times 7 !}+\frac{10 !}{4 ! \times 6 !}+\frac{10 !}{5 ! \times 5 !}+\frac{10 !}{6 ! \times 4 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 1 \times 7 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(+\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(=120+210+252+210=792\)

Explanation:

E Given,
\(\mathrm{n}(\mathrm{A})=4 \mathrm{n}(\mathrm{B})=3\)
\(\mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=24\)
\(\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B}) \times \mathrm{n}(\mathrm{C})\)
\(24=4 \times 3 \times \mathrm{n}(\mathrm{c})\)
\(\mathrm{n}(\mathrm{c})=\frac{24}{4 \times 3}=2\)

Explanation:

E Given, \(\mathrm{n}(\mathrm{A})=5\)
\(\mathrm{n}(\mathrm{B})=7\)
\(\therefore\) Number of relation on
\(A \times B=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]}\)
\(\mathrm{A} \times \mathrm{B}=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]^2}\)
\(2^{(5 \times 7)^2}=2^{35 \times 35}\)

Explanation:

Explanation:

B Given,
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=99\)
Find, \(\quad \mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\) ?
Then,
\(\mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\mathrm{n}\{(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})\}\)
\(=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=99 \times 99=99^2\)

Explanation:

A Number of element in set \(\mathrm{A}=5\)
And no. of element in set \(B=2\)
The no. of element in ordered pair \(A \times B=2 \times 5=10\)
\(\mathrm{n}(\mathrm{A}+\mathrm{B})=10\)
Then, The number of subsets of \(A \times B\) each having at least 3 and at most 6 elements is-
\(={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6\)
We know that,
\({ }^n c_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{10 !}{3 ! \times 7 !}+\frac{10 !}{4 ! \times 6 !}+\frac{10 !}{5 ! \times 5 !}+\frac{10 !}{6 ! \times 4 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 1 \times 7 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(+\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(=120+210+252+210=792\)

Explanation:

E Given,
\(\mathrm{n}(\mathrm{A})=4 \mathrm{n}(\mathrm{B})=3\)
\(\mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=24\)
\(\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B}) \times \mathrm{n}(\mathrm{C})\)
\(24=4 \times 3 \times \mathrm{n}(\mathrm{c})\)
\(\mathrm{n}(\mathrm{c})=\frac{24}{4 \times 3}=2\)

Explanation:

E Given, \(\mathrm{n}(\mathrm{A})=5\)
\(\mathrm{n}(\mathrm{B})=7\)
\(\therefore\) Number of relation on
\(A \times B=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]}\)
\(\mathrm{A} \times \mathrm{B}=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]^2}\)
\(2^{(5 \times 7)^2}=2^{35 \times 35}\)

Explanation:

Explanation:

B Given,
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=99\)
Find, \(\quad \mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\) ?
Then,
\(\mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\mathrm{n}\{(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})\}\)
\(=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=99 \times 99=99^2\)

Explanation:

A Number of element in set \(\mathrm{A}=5\)
And no. of element in set \(B=2\)
The no. of element in ordered pair \(A \times B=2 \times 5=10\)
\(\mathrm{n}(\mathrm{A}+\mathrm{B})=10\)
Then, The number of subsets of \(A \times B\) each having at least 3 and at most 6 elements is-
\(={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6\)
We know that,
\({ }^n c_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{10 !}{3 ! \times 7 !}+\frac{10 !}{4 ! \times 6 !}+\frac{10 !}{5 ! \times 5 !}+\frac{10 !}{6 ! \times 4 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 1 \times 7 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(+\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(=120+210+252+210=792\)

Explanation:

E Given,
\(\mathrm{n}(\mathrm{A})=4 \mathrm{n}(\mathrm{B})=3\)
\(\mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=24\)
\(\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B}) \times \mathrm{n}(\mathrm{C})\)
\(24=4 \times 3 \times \mathrm{n}(\mathrm{c})\)
\(\mathrm{n}(\mathrm{c})=\frac{24}{4 \times 3}=2\)

Explanation:

E Given, \(\mathrm{n}(\mathrm{A})=5\)
\(\mathrm{n}(\mathrm{B})=7\)
\(\therefore\) Number of relation on
\(A \times B=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]}\)
\(\mathrm{A} \times \mathrm{B}=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]^2}\)
\(2^{(5 \times 7)^2}=2^{35 \times 35}\)

Explanation:

Explanation:

B Given,
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=99\)
Find, \(\quad \mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\) ?
Then,
\(\mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\mathrm{n}\{(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})\}\)
\(=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=99 \times 99=99^2\)

Explanation:

A Number of element in set \(\mathrm{A}=5\)
And no. of element in set \(B=2\)
The no. of element in ordered pair \(A \times B=2 \times 5=10\)
\(\mathrm{n}(\mathrm{A}+\mathrm{B})=10\)
Then, The number of subsets of \(A \times B\) each having at least 3 and at most 6 elements is-
\(={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6\)
We know that,
\({ }^n c_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{10 !}{3 ! \times 7 !}+\frac{10 !}{4 ! \times 6 !}+\frac{10 !}{5 ! \times 5 !}+\frac{10 !}{6 ! \times 4 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 1 \times 7 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(+\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(=120+210+252+210=792\)

Explanation:

E Given,
\(\mathrm{n}(\mathrm{A})=4 \mathrm{n}(\mathrm{B})=3\)
\(\mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=24\)
\(\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B}) \times \mathrm{n}(\mathrm{C})\)
\(24=4 \times 3 \times \mathrm{n}(\mathrm{c})\)
\(\mathrm{n}(\mathrm{c})=\frac{24}{4 \times 3}=2\)

Explanation:

E Given, \(\mathrm{n}(\mathrm{A})=5\)
\(\mathrm{n}(\mathrm{B})=7\)
\(\therefore\) Number of relation on
\(A \times B=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]}\)
\(\mathrm{A} \times \mathrm{B}=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]^2}\)
\(2^{(5 \times 7)^2}=2^{35 \times 35}\)

Explanation:

Explanation:

B Given,
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=99\)
Find, \(\quad \mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\) ?
Then,
\(\mathrm{n}\{(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})\}=\mathrm{n}\{(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})\}\)
\(=\mathrm{n}(\mathrm{A} \cap \mathrm{B}) \times \mathrm{n}(\mathrm{B} \cap \mathrm{A})=99 \times 99=99^2\)

Explanation:

A Number of element in set \(\mathrm{A}=5\)
And no. of element in set \(B=2\)
The no. of element in ordered pair \(A \times B=2 \times 5=10\)
\(\mathrm{n}(\mathrm{A}+\mathrm{B})=10\)
Then, The number of subsets of \(A \times B\) each having at least 3 and at most 6 elements is-
\(={ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5+{ }^{10} \mathrm{C}_6\)
We know that,
\({ }^n c_r=\frac{n !}{r !(n-r) !}\)
\(=\frac{10 !}{3 ! \times 7 !}+\frac{10 !}{4 ! \times 6 !}+\frac{10 !}{5 ! \times 5 !}+\frac{10 !}{6 ! \times 4 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 1 \times 7 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(+\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !}+\frac{10 \times 9 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !}\)
\(=120+210+252+210=792\)

Explanation:

E Given,
\(\mathrm{n}(\mathrm{A})=4 \mathrm{n}(\mathrm{B})=3\)
\(\mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=24\)
\(\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B} \times \mathrm{C})=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B}) \times \mathrm{n}(\mathrm{C})\)
\(24=4 \times 3 \times \mathrm{n}(\mathrm{c})\)
\(\mathrm{n}(\mathrm{c})=\frac{24}{4 \times 3}=2\)

Explanation:

E Given, \(\mathrm{n}(\mathrm{A})=5\)
\(\mathrm{n}(\mathrm{B})=7\)
\(\therefore\) Number of relation on
\(A \times B=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]}\)
\(\mathrm{A} \times \mathrm{B}=2^{[\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})]^2}\)
\(2^{(5 \times 7)^2}=2^{35 \times 35}\)