MCQ
In how many ways can 12 people be divided into 3 groups where 4 persons must be there in each group?
- A$\text{None of these}$
- B$\frac{12!}{(4!)^3}$
- C$\text{Insufficient data}$
- D$\frac{12!}{3!\times(4!)^3}$
Solution:
Number of ways in which
$\text{m}\times\text{n"}>$
$\text{m}\times\text{n}$ distinct things can be divided equally into n
$\text{n"}>$ groups
$=\frac{(\text{mn})!}{\text{n}!\times(\text{m}!)\text{n}}$
Given, $12(3\times4)$ people needs to be divided into 3 groups where 4 persons must be there in each group.
So, the required number of ways $=\frac{({12})!}{{3}!\times(4!)\text{n}}$
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The minimum value of $4^{\text{x}}+4^{1-\text{x}},\text{x}\in\text{R}$ is: