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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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Fix the kingdoms and permute 2n champions$\implies (2n)!\; ways$

In this  $(2n)!$  every kingdom counted its champions twice. i.e.  $(C_i,C_j)$  and  $(C_j,C_i)$  both are counted..

So we should have to divide (2n)! by 2 for each kingdom. And there are n kingdoms.

$\implies \frac{(2n)!}{2^n}$

 

 

Answer:

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