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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?

# 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}
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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?

# 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....

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}

7. {AD}{BC}{EF}
8. {AD}{BE}{CF}
9. {AD}{BF}{CE}

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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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.

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

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.

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

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.

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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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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.
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