Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

6 people form groups of 2 for a practical work. Each group [#permalink]
05 Jun 2010, 14:53

1

This post received KUDOS

7

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

33% (01:56) correct
67% (00:46) wrong based on 10 sessions

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?

Re: Combinations problems [#permalink]
05 Jun 2010, 15:21

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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: \(C^2_6*C^2_4*C^2_2=90\).

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.

Re: Combinations problems [#permalink]
07 Jun 2010, 05:35

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: .

Answer: 90.

Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate....

Re: Combinations problems [#permalink]
07 Jun 2010, 05:53

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: \(C^2_6*C^2_4*C^2_2=90\).

Re: Combinations problems [#permalink]
07 Jun 2010, 07:23

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

bibha 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?

# of ways 6 people can be divided into 3 groups when order matters is: .

Answer: 90.

Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate....

nitishmahajan wrote:

Bunuel,

Dont you think we need to multiply 90 with 3! as we have 3 teams now and 3 teams needs to be given one place amongst asia, europe or Africa,

Please correct me if my understanding is wrong here.

There are 15 ways 6 people can be divided equally into 3 groups, each containing 2 persons when the order of the groups is not important (meaning that we don't have group #1, group #2, and group #3):

We can get this # from the following formula: \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\). We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important.

Next, we are told that each group is assigned to one of three continents: Asia, Europe or Africa. So now the order of the groups IS important as we can assign group #1 to Asia, #2 to Europe and #3 to Africa OR #1 to Europe, #2 to Asia and #3 to Africa ... Several different assignments are possible. So how many?

Let's consider division #1 (1. {AB}{CD}{EF}), in how many ways can we assign these groups to the given 3 countries?

Total 6 different assignments, basically the # of permutations of 3 distinct objects (3!): {AB}, {CD}, and {EF}.

3!=6 different assignments for one particular division, means that for 15 divisions there will be total of 3!*15=90 assignments possible.

So when the the order of the groups is not important we are dividing \(C^2_6*C^2_4*C^2_2\) by the factorial of the # of groups - 3! --> \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\);

But when the order of the groups is important (when we are assigning them to certain task) divisionj is not deeded: \(C^2_6*C^2_4*C^2_2=90\).

Re: Combinations problems [#permalink]
07 Jun 2010, 07:28

Bunuel wrote:

bibha 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?

# of ways 6 people can be divided into 3 groups when order matters is: .

Answer: 90.

Bunuel, I have always found your explanations brilliant. With this question, however, i could not grasp your explanation. Please kindly elaborate....

nitishmahajan wrote:

Bunuel,

Dont you think we need to multiply 90 with 3! as we have 3 teams now and 3 teams needs to be given one place amongst asia, europe or Africa,

Please correct me if my understanding is wrong here.

There are 15 ways 6 people can be divided equally into 3 groups, each containing 2 persons when the order of the groups is not important (meaning that we don't have group #1, group #2, and group #3):

We can get this # from the following formula: \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\). We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important.

Next, we are told that each group is assigned to one of three continents: Asia, Europe or Africa. So now the order of the groups IS important as we can assign group #1 to Asia, #2 to Europe and #3 to Africa OR #1 to Europe, #2 to Asia and #3 to Africa ... Several different assignments are possible. So how many?

Let's consider division #1 (1. {AB}{CD}{EF}), in how many ways can we assign these groups to the given 3 countries?

Total 6 different assignments, basically the # of permutations of 3 distinct objects (3!): {AB}, {CD}, and {EF}.

3!=6 different assignments for one particular division, means that for 15 divisions there will be total of 3!*15=90 assignments possible.

So when the the order of the groups is not important we are dividing \(C^2_6*C^2_4*C^2_2\) by the factorial of the # of groups - 3! --> \(\frac{C^2_6*C^2_4*C^2_2}{3!}=15\);

But when the order of the groups is important (when we are assigning them to certain task) divisionj is not deeded: \(C^2_6*C^2_4*C^2_2=90\).

Re: Combinations problems [#permalink]
21 Jun 2010, 06:35

1

This post received KUDOS

Expert's post

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?

Thanks for your help.

Yes I calculated the probability instead of # of groups. Though the approach would be exactly the same: {total # of groups}-{# of groups with only women} = {# of groups with at least one man} --> \(C^3_{10}-C^3_6=120-20=100\). Or as you wrote 5/6th of total # 120 = 100. _________________

Re: Combinations problems [#permalink]
24 Jun 2010, 04:34

Expert's post

kobra92 wrote:

buneul, just one more favor....

please let me know if my approach is correct.

first i calculated the number of ways in which at least one man is chosen

1. MWW 4 2. MMW 6 3 MMM 4

then for each i calculated the number of ways the women can be filled in... so for 1. 6!/2!(4!) = 15 2. 6!/5! = 6 3. none

then 4(15) + 6(6) + 4 = 100

You are basically calculating # of groups with at least 1 man in direct way, which is longer than the approach "all minus none" but still is a correct solution.

Direct way: \(C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=4*15+6*6+4=100\). _________________

Re: Combinations problems [#permalink]
23 Oct 2010, 19:11

The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important - why divide? The answer is right, but I cant understand how we arrived at the formula/division.

Re: Combinations problems [#permalink]
24 Oct 2010, 02:50

Expert's post

vicksikand wrote:

The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important - why divide? The answer is right, but I cant understand how we arrived at the formula/division.

Re: Combinations problems [#permalink]
24 Oct 2010, 04:25

Bunuel wrote:

vicksikand wrote:

The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important - why divide? The answer is right, but I cant understand how we arrived at the formula/division.

Thats the explanation I was talking about. I read it - it totally makes sense , but how did you arrive at the formula? "We can get this # from the following formula: \frac{C^2_6*C^2_4*C^2_2}{3!}=15. We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important." nCr is used for selections when order isnt important, nPr when order is important.

Re: Combinations problems [#permalink]
24 Oct 2010, 04:28

1

This post received KUDOS

Expert's post

vicksikand wrote:

Bunuel wrote:

vicksikand wrote:

The explanation is clear, but I have a few doubts: 1. We define nCr as r selections out of n when the order is not important. If the order is important we use nPr. 2. 6C2 (2 out of 6)x 4C2 (2 out of 4) x 2C2(2 out of 2), now since "C" by definition is used for selections when order is not important - why divide? The answer is right, but I cant understand how we arrived at the formula/division.

Thats the explanation I was talking about. I read it - it totally makes sense , but how did you arrive at the formula? "We can get this # from the following formula: \frac{C^2_6*C^2_4*C^2_2}{3!}=15. We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important." nCr is used for selections when order isnt important, nPr when order is important.

Re: Combinations problems [#permalink]
24 Oct 2010, 04:43

Hope it helps.[/quote] Thats the explanation I was talking about. I read it - it totally makes sense , but how did you arrive at the formula? "We can get this # from the following formula: \frac{C^2_6*C^2_4*C^2_2}{3!}=15. We are dividing by 3! (factorial of the # of groups) because the order of the groups is not important." nCr is used for selections when order isnt important, nPr when order is important.[/quote]

Re: Combinations problems [#permalink]
31 Oct 2010, 08:25

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.

Bunuel, why does order matter in the question above?

In the question below, the answer would be 2520 if order matters, but the answer is 105. Are these two questions not essentially the same?

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

Re: Combinations problems [#permalink]
31 Oct 2010, 08:40

Expert's post

Yalephd 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.

Bunuel, why does order matter in the question above?

In the question below, the answer would be 2520 if order matters, but the answer is 105. Are these two questions not essentially the same?

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

Whole above discussion is about this issue. Please read it and follow the links provided in my earlier posts for other examples and discussions of this concept.

But again:

Assignment: Asia - group#1, Europe - group#2, Africa - group#3 is different from Asia - group#3, Europe - group#1, Africa - group#2. That's why the order of the groups matters here.

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 can be sent to 3 different location in 3! ways.

Kindly correct me if i am wrong.

Waiting for your reply. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Although I have taken many lessons from Field Foundations that can be leveraged later, the lessons that will stick with me the strongest have been the emotional intelligence lessons...

Tick, tock, tick...the countdown to January 7, 2016 when orientation week kicks off. Been a tiring but rewarding journey so far and I really can’t wait to...