TIFR CSE 2013 | Part A | Question: 9

Question

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

• A. $\frac{\left ( 2n \right )!}{2^{n}}$
• B. $\frac{\left ( 2n \right )!}{n!}$
• C. $\frac{\left ( 2n \right )!}{2^{n} . n!}$
• D. $\frac{n!}{2}$
• E. None of the above

Comments

Suppose 4 kingdoms and 8 champions, now champions can be selected as follows:
$=C(8,2)*C(6,2)*C(4,2)*C(2,2)$

$=\frac{8!}{2!*6!}*\frac{6!}{2!*4!}*\frac{4!}{2!*2!}*\frac{2!}{2!*0!}$

If we cancel the terms in numerator and denominator then we will be left with:

 $=\frac{8!}{2!*2!*2!*2!} = \frac{(2n)!}{2^n}$

Hence, option (A) is correct!

PS: it is assumed that all the kingdoms are different, if not, then we will divide by n!.

Answer 1

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.

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