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05 Jun 2010, 15:53
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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?
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?
07 Jun 2010, 08:23
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bibha
nitishmahajan
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.
05 Jun 2010, 16:21
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Gusano97
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: \(\frac{5}{6}\)
