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m06Q11 [#permalink] New post 04 Jul 2009, 06:37
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Question Stats:

28% (02:09) correct 72% (01:25) wrong based on 204 sessions
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

(A) 90
(B) 105
(C) 168
(D) 420
(E) 2520

[Reveal] Spoiler: OA
B

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pl explain the ans particularly usage of 4!.
[Reveal] Spoiler: OA
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Re: m06Q11 [#permalink] New post 04 Jul 2009, 14:10
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May be someone with a good understanding of permutations and combinations can explain this discrepency.
Personally I never doubt the explanations given by AkamaiBrah but i dont know about this particular question.
May be mods can help.
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Re: m06Q11 [#permalink] New post 05 Jul 2009, 01:54
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vaivish1723 wrote:
pl explain the ans particularly usage of 4!.


By multiplying all the combinations of different groups you are getting a figure for which order matters

For example if there are 8 people: john,mary,jane,rima,sandeep,huhu,bruce and ram

the groups (john,mary)(jane,rima)(sandeep,huhu)(bruce,ram) is the same as (bruce,ram)(john,mary)(sandeep,huhu)(jane,rima)

as mentioned earlier by multiplying the number of various combinations possible you get all the different ways the groups could be organised without differentiating that order does not patter. The way to discount this is to divide by the factorial of the number of items = 4!

I hope that makes sense

((28)(15)(6)(1))/(4)(3)(2) = 105.
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Re: m06Q11 [#permalink] New post 25 Oct 2010, 04:26
Could you please explain how you came up with each number: ((28)(15)(6)(1))/(4)(3)(2) = 105.

I am confused. Would be really helpful.
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Re: m06Q11 [#permalink] New post 25 Oct 2010, 05:02
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ileannaconst wrote:
Could you please explain how you came up with each number: ((28)(15)(6)(1))/(4)(3)(2) = 105.

I am confused. Would be really helpful.

get 2 people from 8 = \frac{8!}{6!*2!}=28
get 2 people from 8-2 = \frac{6!}{4!*2!}=15
get 2 people from 8-4 = \frac{4!}{2!*2!}=6
get 2 people from 8-6 = \frac{2!}{0!*2!}=1
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Re: m06Q11 [#permalink] New post 25 Oct 2010, 05:11
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Hi,
I didn't know that we had to divide by 4!, I just knew that we had to divide by some number to make up for the repetitions. So I didn't know what to do with the 840 I kept on getting.
So I left PnC and just tried another logic:

Here it is..

I took up a smaller group of 4 persons, whom I have to divide into 2 teams - A B C D

So the 1st team could have any one of AB, AC or AD - 3 and the other team in each of the cases has only one option - the other 2 persons.
-- So it becomes 3 x 1
When I extended it to 6 persons into 2 teams, I got - 5 x 3 x 1

So I extended till 8 and got it 7 x 5 x 3 x 1 = 105 :P
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Re: m06Q11 [#permalink] New post 25 Oct 2010, 09:37
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Here is another conceptual way:

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed.
Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed.
Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed.
Step4: Last two can be teamed up in only 1 way.

So total ways = 7*5*3*1 = 105
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Re: m06Q11 [#permalink] New post 15 Jun 2011, 07:46
if we dont go by the brilliant method suggested by srijaidev and instead go with conventional combinatorics as suggested by nightwing:


WHAT SHOULD BE THE WORDING OF THE QUESTION SO THAT WE DO NOT NEED TO DIVIDE THE TOTAL NUMBER OF COMBINATIONS BY 4! ?

I am posting this query so that I can understand the difference in the 2 question types if faced by either on G-day. :?:
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Re: m06Q11 [#permalink] New post 04 Nov 2011, 00:39
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In!/Out!

8!/(4!2!2!2!2!)
= 7 * 5 * 3
= 105
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Re: m06Q11 [#permalink] New post 26 Mar 2012, 05:58
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\frac{1}{4!}*[\frac{8*7}{2!}*\frac{6*5}{2!}*\frac{4*3}{2!}*\frac{2*1}{2!}] = 105

\frac{1}{4!} because order of groups does not matter.
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Re: m06Q11 [#permalink] New post 26 Mar 2012, 06:20
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vaibhavtripathi wrote:
Here is another conceptual way:

Step1: Lets take one person from 8. Now in how many ways can he be teamed up with another man? 7!!! Team-1 is formed.
Step2: Lets take another person from remaining 6. In how many ways can he be teamed up with another man? 5!!! Team-2 is formed.
Step3: Lets take another person from remaining 4. In how many ways can he be teamed up with another man? 3!!! Team-3 is formed.
Step4: Last two can be teamed up in only 1 way.

So total ways = 7*5*3*1 = 105


For poor chaps like me who find the permutation, combination, probablity as a demon, your answers were simple to understand buddy....

+1 to u
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Re: m06Q11 [#permalink] New post 26 Mar 2012, 06:46
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vaivish1723 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

(A) 90
(B) 105
(C) 168
(D) 420
(E) 2520

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

pl explain the ans particularly usage of 4!.


\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

Answer: B.

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups

There is also direct formula for this:

1. 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*m!}.

2. 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}

Hope it helps.
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Re: m06Q11 [#permalink] New post 07 Jun 2012, 13:59
Hi,

I still have some trouble to figure out when to account for repetitions :x . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy :?

Thank you so much

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Re: m06Q11 [#permalink] New post 07 Jun 2012, 21:59
What's the probability that combinatorics appear on the exam?
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Re: m06Q11 [#permalink] New post 08 Jun 2012, 02:47
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Elzi wrote:
Hi,

I still have some trouble to figure out when to account for repetitions :x . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy :?

Thank you so much

Elzi


This concept is explained here:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups
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Re: m06Q11 [#permalink] New post 08 Jun 2012, 14:42
Thanks a lot Bunuel!
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Re: m06Q11 [#permalink] New post 29 Oct 2012, 10:01
Another option is to use the formula (found it in one of the threads):

(m*n)!/[(n!)^m]*m!
m - number of people
n - number of groups

8!/[(2)^4]*4! = 105
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Re: m06Q11 [#permalink] New post 29 Oct 2012, 10:06
Expert's post
EV wrote:
Another option is to use the formula (found it in one of the threads):

(m*n)!/[(n!)^m]*m!
m - number of people
n - number of groups

8!/[(2)^4]*4! = 105


That's correct. Check here for the formula: m06q11-80504.html#p1065072
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Re: m06Q11 [#permalink] New post 03 Nov 2012, 00:44
Simplify the question in this way, you need to select a team of two from 8 = 8c2 = 28
now 6 are left and = 6c2 = 15 4c2 = 6 and finally only two people left =1

But you can select these in 4! ways.
so final selection will be (8c2 * 6c2 * 4c2 * 2c2) / (4 !) = 105
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Re: m06Q11 [#permalink] New post 25 Oct 2013, 23:55
Elzi wrote:
Hi,

I still have some trouble to figure out when to account for repetitions :x . I thought that by calculating the number of possibilities for the first, second, third and fourth team, I create 4 different and independent "pools" of choices.
To me, the problem seemed similar to the following:

9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
(OA is 60 - first step: calculate possibilities for the two different pools - second step: multiply by each other WITHOUT accounting for repetitions.

It would be great if anyone could help me on this, it drives me crazy :?

Thank you so much

Elzi


I think the trick is not to get too technical, but to rely on logic. Here's how I did that (and feel free to correct my thinking)

-I have 8 people and I need to figure out how many DIFFERENT ways I can split these guys into 4 teams of 2. I'm making a chain of 4 teams. A chain is different only when there's a new team in there, not a different arrangement of the same 4 teams!
-Let me call these teams T1, T2, T3, T4
-Clearly what the question is saying is that I don't care if I pick them the following ways, it's all the same to the author
T2 T1 T3 T4
T2 T3 T1 T4
and 22 other varieties (i.e. a total of 4! or 24 ways I can arrange these 4 SPECIFIC teams) - ok, noted
-Let's see how I can choose T1, T2, T3, T4
T1 - I can choose 8C2 ways
T2 - I can choose 6C2 ways
T3 - I can choose 4C2 ways
T4 - I can choose 2C2 or 1 way (since well there are only 2 guys left)

The key is NOT to get confused on the usage of combinations and assume that since we're using combinations the order already doesnt matter. Sure in the last team say only Jack and Jill were left - we still only picked one team - i.e. we didn't distinguish between (Jack-Jill) or (Jill-Jack). This doesnt mean that we are done. We need to focus on what the question is asking us and we need to give them DISTINCT ways of PICKING 4 TEAMS of 2 (i.e. this is one further layer)

8C2 * 6C2 * 4C2 * 2C2 = total number of ways I can arrange Teams 1, 2, 3 and 4. But I don't care about how these teams are arranged. Therefore, I divide by 4!

It helps to try and understand what the author is asking rather than get stuck on what formula to use.

I'm also trying to solidify my understanding of when to adjust for arrangements etc. so this is work-in-progress!

Hope this helps.
Re: m06Q11   [#permalink] 25 Oct 2013, 23:55
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