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22 votes
22 votes

There are $n$ kingdoms and $2n$ champions. Each kingdom gets $2$ champions. The number of ways in which this can be done is:

  1. $\frac{\left ( 2n \right )!}{2^{n}}$
  2. $\frac{\left ( 2n \right )!}{n!}$
  3. $\frac{\left ( 2n \right )!}{2^{n} . n!}$
  4. $\frac{n!}{2}$
  5. None of the above
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5 Answers

Best answer
37 votes
37 votes

We have $n$ Kingdoms as ${k_1}, {k_2},\ldots , k_n.$

Firstly we can select $2$ champions from ${2n}$ champions and assign to ${k_1}= \binom{2n}{2}$ ways (Say ${w_1}$)

Then we can select next $2$ champions and assign to ${k_2} = \binom{2n-2}{2}$ ways (Say ${w_2}$)

and so on..

For last kingdom , we have $2$ champions left$ = \binom{2}{2}$ ways (Say ${w_n}$)

$\begin{align}\text{Total ways for assigning $2n$ champions to $n$ kingdoms} &= {w_1} * {w_2} *\ldots *{w_n} \\ &= \binom{2n}{2} * \binom{2n - 2}{2} * . . . * \binom{2}{2} \\&=\frac{(2n)!}{2^{n}} \end{align}$ 

So, Option A  (Ans) . 

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9 votes
9 votes

We can do this question by simple assuming value :-

Suppose n=2 it means there is 2 kingdoms and 4 champions (A,B,C,D)

Put n=2 in

option A :-4!/4 = 6

Option B:-4!/2! = 12

Option C:-4!/4*2!= 3 

Option D:-4!/2 = 12 

So option A is correct option 

6 votes
6 votes

n Kingdoms and 2n champions and each kingdom must have 2 champions each.

This is similar to distributing n distinct objects into n distinct boxes where each box must have a certain number of objects.

The formula applicable is shown by the above theorem.

Hence, No. of ways = (2n)!/(2!)n = (2n)!/2n 

Answer- A

2 votes
2 votes
2n Champion to be distributed among n kingdom with 2 members each if we shuffle members then two kingdom will get different champion . So, no. of way

=$\frac{(2n)!}{\underbrace{2!.2!.2! \dots 2!}_{n \text{ times }} .n!} $
=$\frac{(2n)!}{2^n . n!}$

Option C
Answer:

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