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):
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+groupscombination-55369.html#p690842probability-88685.html#p669025combination-groups-and-that-stuff-85707.html#p642634sub-committee-86346.html?highlight=divide+groupsHope it's clear.