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Director  Joined: 22 Mar 2011
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WE: Science (Education)

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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: $$C^2_6*C^2_4*C^2_2=90$$.

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.

You are right, it should be 90*3! = 540. Order of groups matters here, as we have different continents.
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Senior Manager  Joined: 13 Aug 2012
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Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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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.

Senior Manager  Joined: 13 Aug 2012
Posts: 391
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Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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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 $$=\frac{4!}{2!2!}*\frac{6!}{1!5!} = 6 * 6 = 36$$

$$=60+4+36 = 100$$

Manager  B
Joined: 23 Sep 2013
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6 people form groups of 2 for a practical work. Each group  [#permalink]

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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: $$C^2_6*C^2_4*C^2_2=90$$.

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 .

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?

Regards
SR
Manager  B
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Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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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?

For question B) what is wrong with the following approach?

We need at least one man: _ _ _. So lets choose one man: 4C1. Rest of the two places can be occupied by anyone since we have fulfilled the requirement of at least one man. So, that can be done in 9C2 ways. Total = 4C1*9C2 = 144 ways.
Manager  G
Joined: 27 Jan 2016
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Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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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?

1) 6c2 * 4c2 * 2c2

2)10c3-6c3
Manager  G
Joined: 27 Jan 2016
Posts: 122
Schools: ISB '18
GMAT 1: 700 Q50 V34
Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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capicu006 wrote:
Bunuel,

I'm confused on the second question. Doesn't it ask for a specific number, not a probability?

In any event, would it by 5/6ths of the total number --> 5/6ths of 120 --> 100?

You are right.
The solution is 10c3-6c3
Intern  B
Joined: 14 Apr 2018
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Location: India
Schools: NUS '20
6 people form groups of 2 for a practical work. Each group  [#permalink]

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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: $$C^2_6*C^2_4*C^2_2=90$$.

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.

I also have the same doubt. Please explain?
Intern  Joined: 01 Apr 2020
Posts: 6
Re: 6 people form groups of 2 for a practical work. Each group  [#permalink]

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Q. 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?

A. We can also use a formula of [(mn)! / {(n!)^m x m!}] where (m x n) items, divided in m groups of n objects each where orders does not matter. Multiple with m! if order matters.

In this case m x n items are 6
m = 3 = no of groups
n = 2 = no of objects/people in each group

=> 6!/ 2!^3 x 3! = 15
Since order matters x by m! = 6

=> total number of ways = 90 Re: 6 people form groups of 2 for a practical work. Each group   [#permalink] 02 Apr 2020, 03:55

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