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

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6 people form groups of 2 for a practical work. Each group [#permalink]  05 Jun 2010, 14:53
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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?
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Re: Combinations problems [#permalink]  05 Jun 2010, 15:21
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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?

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: $$\frac{5}{6}$$
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Re: Combinations problems [#permalink]  05 Jun 2010, 15:25
Thank you Bunuel!.
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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: .

Bunuel,
I have always found your explanations brilliant. With this question, however, i could not grasp your explanation.
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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: $$C^2_6*C^2_4*C^2_2=90$$.

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

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

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

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):

1. {AB}{CD}{EF}
2. {AB}{CE}{DF}
3. {AB}{CF}{DE}

4. {AC}{BD}{EF}
5. {AC}{BE}{DF}
6. {AC}{BF}{DE}

10. {AE}{CD}{BF}
11. {AE}{BC}{DF}
12. {AE}{CF}{BD}

13. {AF}{CD}{BE}
14. {AF}{BC}{DE}
15. {AF}{CE}{BD}

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?

Asia - Europe - Africa
{AB} - {CD} - {EF}
{AB} - {EF} - {CD}
{CD} - {AB} - {EF}
{CD} - {EF} - {AB}
{EF} - {AB} - {CD}
{EF} - {CD} - {AB}

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

Pleas check the following links for similar problems:
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

Hope it's clear.
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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: .

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

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):

1. {AB}{CD}{EF}
2. {AB}{CE}{DF}
3. {AB}{CF}{DE}

4. {AC}{BD}{EF}
5. {AC}{BE}{DF}
6. {AC}{BF}{DE}

10. {AE}{CD}{BF}
11. {AE}{BC}{DF}
12. {AE}{CF}{BD}

13. {AF}{CD}{BE}
14. {AF}{BC}{DE}
15. {AF}{CE}{BD}

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?

Asia - Europe - Africa
{AB} - {CD} - {EF}
{AB} - {EF} - {CD}
{CD} - {AB} - {EF}
{CD} - {EF} - {AB}
{EF} - {AB} - {CD}
{EF} - {CD} - {AB}

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

Pleas check the following links for similar problems:
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

Hope it's clear.

Thanks Bunuel, Its clear now ..!
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Re: Combinations problems [#permalink]  21 Jun 2010, 06:24
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?

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

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.
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Re: Combinations problems [#permalink]  23 Jun 2010, 22:56
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
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Kudos [?]: 47409 [0], given: 7123

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: $$C^1_4*C^2_6+C^2_4*C^1_6+C^3_4=4*15+6*6+4=100$$.
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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.
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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.

Similar problems about the same concept:
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

Hope it helps.
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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.

Similar problems about the same concept:
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

Hope it helps.

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.
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Posts: 28781
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Re: Combinations problems [#permalink]  24 Oct 2010, 04:28
1
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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.

Similar problems about the same concept:
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

Hope it helps.

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.

combination-55369.html#p690685
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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]

combination-55369.html#p690685[/quote]
clear ! Thanks
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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: $$C^2_6*C^2_4*C^2_2=90$$.

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
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Kudos [?]: 47409 [0], given: 7123

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

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

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.

Second problem discussed here: combination-groups-and-that-stuff-85707.html#p642634

In this problem there is no group#1, group#2, group#3, group#4 so 2520 should be divided by 4!=105 to get rid of duplications.
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Re: Combinations problems [#permalink]  16 Sep 2012, 06:10
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 can be sent to 3 different location in 3! ways.

Kindly correct me if i am wrong.

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Re: Combinations problems [#permalink]  16 Sep 2012, 06:10
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.

_________________

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

Re: Combinations problems   [#permalink] 16 Sep 2012, 06:10

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