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

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) . 

Comments

@Ayush Upadhyaya I have some doubt in this question like:

1. If we consider this question https://gateoverflow.in/21004/tifr2012-a-7 it is very much similar to this question but it is mentioned in the previous question that each team is disjoint but in this question it is not mentioned but in the answer disjoint is taken. So why the answer shouldn't be 2nC2 * 2nC2....

Sir please clear this doubt. 

---

@rajatmyname-The difference comes due to the fact that here the distribution is ordered and in the question link you have given, distribution is unordered(Teams are not labelled).

---

Ayush Upadhyaya Thanks a lot for explaining the difference b/w the two.

Answer 2

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 

Answer 3

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

Answer 4

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

Comments

Yes sir , that is what my question/doubt is.We know that knigdom cannot be identical , but in the question nothing is mentioned.So in that case we will take it as normal meaning.?

Becuase if the question would be unlabelled kingdom then answer will be $C$  else answer will be $A$

---

"normal meaning" -- yes, we should always use that. Like when we talk about people -- no 2 can be identical. Only for things which can be duplicated, repetition comes. Applying "normal meaning" is quite handy for answering many GATE questions -- 'A' in GATE is for "Aptitude".

---

thanks a lot, sir :)