Permutations & Cobination question

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GMAT 2009

Permutations & Cobination question (Question)

Harikishore said – Tue, 23 Jun 2009 06:29:58 -0000 ( Edit Link )

Hi! Please solve this problem: In howmany ways a 2 membered 4 teams can be formed from a group of 8 people? While solving please explain, why that pirticular formula or argument is used. I am finding that mere knowing of formulae is not helping as one need to identify first the category under which the problem falls such as is it a permuatation or combination problem or repetition allowed/not allowed etc, . Can any body broadly classify various categories exists in permutation combination problems, so that it becomes easy to categorise the problem first and then apply suitable formula/argument

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fdsa1 said – Tue, 23 Jun 2009 13:29:39 -0000 ( Flag Edit Mark / Unmark Best Link )

I think the answer is 168. (Please let me know if it is not correct).

Explanation: Permutation problem is the one that deals with the very detail of the problem. It deals with how can the members be arranged among themselves. For eg., if you were given that out of group of 8 people(Rita, Sita, Ram, Shyam… and four others), in how many ways can teams be formed such that Rita and Sita belong to the same team – in this case it is permutation. In here, it is different. I think it is a combination problem as we dont have to find out anything according to the order, if Sita was selected first or Rita was second one… So, considering it to be a combination problem, I guess the solution is C(8,4) – 4 teams out of 8 people and C(4,2) – 2 members in each team. So, the solution will be C(8,4) * C(4,2).

Thanks!
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Harikishore said – Wed, 24 Jun 2009 05:05:34 -0000 ( Flag Edit Mark / Unmark Best Link )

Thank you fdsa1 for a detailed explanation. But C(8,4)* C(4,2) is 420 (70×6).

The solution could be….. Let us say that we choose 4 people out of 8 as one memeber of each team. This can be done in C(8,4) ways. Now we are left with 4 slots by the side of each member selected and we have 4 people to select from. This can be filled in C(4,4) ways . Hence total number of ways are C(8,4) * C(4,4) i.e 70 ways. This is a pure combination problem since the order of the pair (team) has no relevance as team (1,2) is same as team (2,1).

This is what i think could be the solution. Please correct me If I am wrong.
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fdsa1 said – Wed, 24 Jun 2009 12:34:36 -0000 ( Flag Edit Mark / Unmark Best Link )

C(8,4) * C(4,2) = 168 is correct combination calculation – not sure if it is the right answer or not. I think as you are saying the correct answer to the problem is C(8,4) * C(4,4) = 14. C(n,k)= n.k!. So, the answer would be 14, which is too less for 8 members. So, I am not sure what is the actual answer.
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Sureshbala said – Wed, 24 Jun 2009 19:07:07 -0000 ( Flag Edit Mark / Unmark Best Link )

Hi,

Here is the solution.

The question is all about choosing 2 people each time.

So it has to be
$^8C_2 \times ^6C_2 \times ^4C_2 \times ^2C_2$

$= \frac{8!}{2! \times 6!} \times \frac{6!}{2! \times 4!} \times \frac{4!}{2! \times 2!}$

$= \frac{8!}{(2!)^4}$

Since the order of arrangement is not important here, we have to divide this by 4!.

Hence the answer is
$\frac{8!}{4! \times (2!)^4}= 70$
The same thing is given in the following formulae

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is
$[\frac{(mn)!}{(n!) ^m}] \frac{1}{m!}$
The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is
$\frac{(mn)!}{(n!) ^m}$
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1. Carsten said – Thu, 25 Jun 2009 21:58:35 -0000 ( Flag Edit Link )
  
  Great response!
  
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Harikishore said – Thu, 25 Jun 2009 04:03:36 -0000 ( Flag Edit Mark / Unmark Best Link )
Thank you Sureshbala! As I said, Identification of correct category is the key. First to see that it is a conbination problem, then apply the formula ” of number of ways in which mn different items can be divided equally into m groups” .
```
Is this a coincidence that my answer is also 70? Or is my methodology  also correct? Just for curiosity sake... pl let me know.
```
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chandra_avinash said – Fri, 26 Jun 2009 12:20:48 -0000 ( Flag Edit Mark / Unmark Best Link )

hi fdsa1:

i feel there is an error in your approach;

8C4 would not be the number of ways in which you could choose 4 teams out of a total of 8 people; rather it would be the number of ways of picking 4 guys out of a total of 8;

the question implies that you need to make 4 teams; hence a straightforward way would be picking 2 people per team and not picking 1 person per team and deciding how you could pick the other 4 people;

taking the same approach ahead, you’d have to pick the other 4 people: it would be 4C4

hence the total number of ways would be (8C4)*(4C4)

we’re not bothered about the order – hence it’s a simple question of picking the right pair; does this make things clear?
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fdsa1 said – Mon, 29 Jun 2009 15:00:09 -0000 ( Flag Edit Mark / Unmark Best Link )

Thanks Chandra_avinash, but I got your explanation…. But, I didn’t quite get if my answer is correct or not. In your solution, you said the answer is (8C4)*(4C4) and I also had the same answer. I thought my approach was wrong, but is the answer correct?

I appreciate your response as that will figure out what my mistake is and help me apply right logic to other problems also.

Thanks!!

-Tide of time sweeps all of us into the future – willingly or unwillingly
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Harikishore said – Thu, 09 Jul 2009 12:15:00 -0000 ( Flag Edit Mark / Unmark Best Link )

Hi Suresh The answer of [8!/(2!)4 * 4!] is 105 not 70. Please correct me if I am wrong.
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1. Sureshbala said – Mon, 13 Jul 2009 01:00:45 -0000 ( Flag Edit Link )
  
  Hi,
  
  You are correct, it has to be 105 and not 70.
  
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rajkrish said – Fri, 10 Jul 2009 10:50:46 -0000 ( Flag Edit Mark / Unmark Best Link )

Hi,

Let the players be 1,2,3,4,5,6,7,8

So the teams can be arranged in C(8,2) which is 28

12|13|14|15|16|17|18| 23|24|25|26|27|28| 34|35|36|37|38| 45|46|47|48| 56|57|58| 67|68| 78|
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1. Sureshbala said – Sat, 18 Jul 2009 04:02:00 -0000 ( Flag Edit Link )
  
  Hi, here you need to select 2 people each 4 times.
  
  But you are selecting only 2 people from the available 8 people.
  
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