Explanation
Assuming, cards are drawn without replacement:
Total possible events $$=^{52}C_2$$
P(both red) $$=\dfrac{ ^{26}C_2}{^{52}C_2}$$
P(both king) $$=\dfrac{^{4}C_2}{^{52}C_2}$$
P(both red & king) $$=\dfrac {^2C_2}{{^{52}C_2}}$$
We use the basic addition rule,
P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)
Required probability $$=\dfrac{^{26}C_2+^{4}C_2-^2C_2}{^{52}C_2}$$
We have total digits $$0,1,2,3,4.$$
We want to form odd number, then at last digit can become $$1$$ or $$3$$.
So, two number are fixed for last.
Now, we have option for $$1st, 2nd, 3rd$$ and $$4th$$ place are $$3, 3, 2, 1$$
So , number of getting odd numbers using given five distinct digit $$=3\times 3\times 2\times 1\times 2=36$$
Hence, this is the answer.
Given that,$$^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$$
Then $$n=?$$
$$ ^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}} $$
$$ \Rightarrow \dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$
$$ \Rightarrow \dfrac{\left( n+1 \right)n!}{3\times 2!\left( n-2 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$
$$ \Rightarrow \dfrac{\left( n+1 \right)}{3}=4 $$
$$ \Rightarrow n+1=12 $$
$$ \Rightarrow n=12-1=11\, $$
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