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6 people form groups of 2 for a practical work. Each group

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Re: Combinations problems [#permalink]

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New post 16 Sep 2012, 09:02
fameatop wrote:
Bunuel wrote:
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?



B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?


Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps.

A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?

# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\).

Answer: 90.

Hi Bunuel,

In reference to the first question, i have a doubt which is- 90 is the no of ways in which 6 people can be divided into 3 groups of 2 persons each.
Shouldn't the answer be 90 x 6 = 540 because these 3 different teams can be sent to 3 different location in 3! ways.

Kindly correct me if i am wrong.

Waiting for your reply.


You are right, it should be 90*3! = 540. Order of groups matters here, as we have different continents.
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]

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New post 28 Dec 2012, 03:35
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?


My approach:
How many ways to select 2-2-2 from 6 people?
\(=\frac{6!}{2!4!}*\frac{4!}{2!2!}*\frac{2!}{2!} = 90\)

How many ways to distribute to 3 groups? \(\frac{3!}{3!}=1\)
We divided by 3! because of 2 2 2 are identical distributions over 3 groups.

Answer: 90

Click here for more details and examples: Distributon on Different Containers
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]

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New post 28 Dec 2012, 03:39
Gusano97 wrote:
A)
In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?


What are our possibilities:
M M M \(=\frac{4!}{3!1!}=4\)
M W W \(=\frac{4!}{1!3!}*\frac{6!}{2!4!}= 4 * 15 = 60\)
M M W \(=\frac{4!}{2!2!}*\frac{6!}{1!5!} = 6 * 6 = 36\)

\(=60+4+36 = 100\)

Answer: 100
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Re: 6 people form groups of 2 for a practical work. Each group [#permalink]

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6 people form groups of 2 for a practical work. Each group [#permalink]

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New post 10 Jun 2016, 05:16
Bunuel wrote:
Gusano97 wrote:
A) 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?



B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?


Hi, and welcome to the Gmat Club. Below are the solutions for your problems. Hope it helps.

A. 6 people form groups of 2 for a practical work. Each group is assigned one of three continents: Asia, Europe or Africa. In how many different ways can the work be organized?

# of ways 6 people can be divided into 3 groups when order matters is: \(C^2_6*C^2_4*C^2_2=90\).

Answer: 90.

Similar topics:
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
probability-88685.html#p669025
combination-groups-and-that-stuff-85707.html#p642634
sub-committee-86346.html?highlight=divide+groups

B. In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?

Let's find the probability of the opposite event and subtract it from 1.

Opposite event would be that in the committee of 3 won't be any man (so only women) - \(P(m=0)=P(w=3)=\frac{C^3_6}{C^3_{10}}=\frac{1}{6}\). \(C^3_6\) - # of ways to choose 3 women out 6 women; \(C^3_{10}\) - total # of ways to choose 3 people out of 10.

\(P(m\geq{1})=1-P(m=0)=1-\frac{1}{6}=\frac{5}{6}\).

Answer: \(\frac{5}{6}\)



Hi Bunuel !


With reference to your one previous post mentioned below:

GENERAL RULE:
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 important is \(\frac{(mn)!}{(n!)^m}\)

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 NOT important is \(\frac{(mn)!}{(n!)^m*m!}\).

Why is order important in both these questions? Perhaps I'm not able to get the real rationale behind 'order'.
I presumed the order to be inconsequential and hence divided the equations in both questions by 2 .

Can you please help me in understanding the concept of order.
Specifically can you please tell how relevance of order create distinct groups(if you can actually mention the groups) in the 2nd example:

B) In a group of 10 people, 6 women and 4 men. If a comission of three people has to be formed with at least one man, how many groups can we form?



Thanks in advance !

Regards
SR
6 people form groups of 2 for a practical work. Each group   [#permalink] 10 Jun 2016, 05:16

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6 people form groups of 2 for a practical work. Each group

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