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It is required to divide the $2n$ members of a club into $n$ disjoint teams of $2$ members each. The teams are not labelled. 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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6 Answers

Best answer
29 votes
29 votes
${2n}$ member to be n teams with $2$ member each and teams are unordered so we can exchange $n$ team member among them.
 
     =$\dfrac{(2n)!}{\underbrace{2!.2!.2! \dots 2!}_{n \text{ times }} \times n!}$
     =$\dfrac{(2n)!}{2^n \times n!}$

Option $C.$
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7 votes
7 votes
there are 2n members in the club so we can select n members in $^{2n}C_n$ ways

while selecting the n members will come in a particular order and arrangement among them is not there

Now we can place the remaining n members with the already selected n members in $n!$ ways

so here we get $^{2n}C_n \times n!$

Now (member 1, member 2) = (member 2, member 1) so we can divide the above equation by 2 n times

final answer is $\frac {^{2n}C_n \times n!}{2\times 2\times 2 \times...n\ times} = \frac{(2n)!}{2^n \times n!} $

 

Answer is $C$
2 votes
2 votes

Take example for 2n = 4

From 4 members select any 2 = 4C2

Remaining 2 members select 2 = 2C2


Combinations possible are :-

  1. (1 2)(3 4)
  2. (1 3)(2 4)
  3. (1 4)(2 3)
  4. (2 3)(1 4)
  5. (2 4)(1 3)
  6. (3 4)(1 2)

In each case there are 2 teams each having 2 members.

As teams are unlabelled; arrangements does not matter.

Therefore in above example team1 = team6 , team2 = team5, team3 = team4. 

Number of teams = 2

Number of arrangements for 2 teams = 2!

Therefore number of ways = (4C2 × 2C2)/2! = 3

So answer is 3 ways when 2n = 4.

Put n = 2 in options and option (c) is satisfied.

Derivation =>

2n members; n teams are possible.

Firstly from 2n select any 2 members = 2nC2 

From remaining 2n-2 select any 2 members = (2n-2)C2 

Similarly go on and and at last select 2 members from 2 = 2C2 

Number of teams = n

Number of arrangements for n teams = n!

So number of ways = (2nC2 × (2n-2)C2 × ........ × 2C2)/n!

 = (2n!) / (2^n × n!)

= option (c)

2 votes
2 votes

We need to divide $2n$ members of the club into $n$ disjoint teams of $2$ members each means, $i \ne j \Longrightarrow \text{Team}_i \cap \text{Team}_j = \phi$ and $|\text{Team}_k| = 2$. 

Let $A = \{a_1,a_2, \dots, a_{2n}\}$ be the members of the club. 

Let $\pi(A)$ be the permutation of $A$, such that total possible permutation be $(2n)!$. Each permutation can be represented as $\pi_i(A)$ for $1 \le i \le 2n$. 

For example, $n=3$

One such $\pi_k(A) = a_3a_6a_1a_4a_5a_2$, then we select first two $a_i$'s and make a team of them, then we select next two $a_i$'s and make a team of them and so on. 

$$\begin{array}{|c|c|c|}
\hline
a_3a_6 & a_1a_4 & a_5a_2\\
\hline
\text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$ 

Now $\pi_j(A) = a_6a_3a_1a_4a_5a_2$,

$$\begin{array}{|c|c|c|}
\hline
a_6a_3 & a_1a_4 & a_5a_2\\
\hline
\text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$ 

Though $\pi_k(A) \neq \pi_j(A)$, yet the teams formed using $\pi_k(A)$ and $\pi_j(A)$ are same!! Since, order of members in the team doesn't matter. Hence, we should divide by $2$ for every $\text{Team}_i$. 

In total there are going to be $(2n)!$ such sequences of $2n$ members and $n$ such teams made on every such sequence, and for every such team we are dividing the total ways by $2$. 

$$(2n)!\frac{1}{2^n}$$

We are also given one more condition to satisfy, i.e. "The teams are not labelled". Until now, we have a working solution for labelled teams, so now lets take $\pi_p(A), \pi_q(A)$ for the above mentioned example to understand this condition.

$\pi_p(A) = a_1a_4a_3a_6a_5a_2$

$\pi_q(A) = a_5a_2a_1a_4a_3a_6$

$$\begin{array}{|c|c|c|c|}
\hline
\pi_k(A) & a_3a_6 & a_1a_4 & a_5a_2\\
\hline
\pi_p(A) & a_1a_4 & a_3a_6 & a_5a_2\\
\hline
\pi_q(A) & a_5a_2 & a_1a_4 & a_3a_6\\
\hline
\vdots & a_5a_2 & a_3a_6 & a_1a_4\\
\hline
\vdots & a_1a_4 & a_5a_2 & a_3a_6\\
\hline
\vdots & a_3a_6 & a_5a_2 & a_1a_4\\
\hline
& \text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$ 

This shows that members partition into team matters and not the team order itself. Above $6$ ways are effectively $1$ way only, hence we need to divide total ways with $n!$ since there are $n$ teams.


Final answer would be 
$$\dfrac{(2n)!}{2^n (n)!}$$ 

 $\textbf{Option (C) is correct}$.

Answer:

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