Hi guys here's another combination type question

Question

I got this from the 1000 math question bank downloaded from this site. Isnt the answer suppose to be 6C2?

6 people are to be divided to 3 groups with the same numbers of people. How many such groups are possible? 90

jaynayak · Answer

Ok we can safely say that there will be 2 persons in each group. 

Now number of ways 2 people can be chosen for the 1st group = 6C2
Number of ways 2 people can be chosen for the second group = 4C2
Number of ways 2 people can be chosen for the 3 group = 2C2

Remember for the second group you are chossing from 4 people not 6 people.

Total = 6C2 * 4C2 *2C2 = 90

kevincan · Answer

Are the groups themselves distingushable? In other words,

Group 1..........Group 2...........Group 3

....AB.................CD..................EF
....EF..................AB..................CD

Are these assignments distinguishable? If not, answer should be 6C2 * 4C2/3!= 6C2= 15

apollo168 · Answer

Hi guys, 

Thanks for being so helpful! I'm really not sure whether the groups are distinguishable though. 

In the event that they are not, can you explain to me the logic of dividing the second term by 3!? 

Thanks

The answer given in the question bank is 90.

ghantark · Answer

Kevin.
Can you pls explaing why the 3!.
Thanks

haas_mba07 · Answer

We can look at this problems as :

1. Form 1 group of 2 people from a group of 6 people

2. Form 1 group of 2 people from a group of 4 people

3. Form 1 group of 2 people from a group of 2 people

(1) = 6x5/2! = 15
(2) = 4x3/2! = 6
(3) = 1

Total number of groups = 15x6x1 = 90

ps_dahiya · Answer

To remove duplicates.

parthac · Answer

which math question bank are you referring to? is it the ps 1000 (a zipped file)? I downloaded that but it does not seem to have answers in the documents. could you please post the file that you are referring to, here again?

Thanks

yezz · Answer

I believe you can think of the problem this way:

a) number of possibilities of forming 3 groups of two is 6C2.

B) each new group has 2! ways of arrangment>

c) they are 3 GROUPS

multiply 3*2! *6C2 = 90

I hope i am right.

old_dream_1976 · Answer

another way of solving is 

6 people 

pick 1 you have 5 choices for the first team (1x5)

for team 2 pick 1 you have 3 choices (1x3)


team three pick 1 you have 1 choice (1x1)

total teams (  with no one serving on more than 1 team  ) = 1x3x5=15

heman · Answer

=6c2.4c2.2c2
=15*6*1
= 90

Heman

Hi guys here's another combination type question

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