Let's take a simple example.
how many different ways can a group of 4 people be divided into 4 groups, with each group containing 2 people?Assume that A.B,C,D are the 4 people
First group of 2 can be formed in 4C2 ways = 4*3/2 = 6
(A,B)
(A,C)
(A,D)
(B,C)
(B,D)
(C,D)
After forming the first group, there are only 2 people remaining ,
So the number of ways second group can be formed is 2C2= 1. That means by default , the remaining 2 people will form the second group.
So 2 groups can be formed in 4C2* 2C2= 6*1 = 6 ways
But we analyze the groups formed ,we can see some repetitions
(A,B) (C,D)(A,C)(B,D)(A,D)(B,C)(B,C)(A,D)(B,D)(A,C)(C,D)(A,B)The no of unique groups formed is only 3. The others groups are just repetitions. (A,B) (C,D) group is counted twice
So no of ways we can form a group of 2 = 4C2* 2C2/2!
We are dividing by 2! is to compensate the repetitions we have counted.
Lets get back to the Question :
In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?No of ways we can form a groups of 3 =
(9C3*6C3*3C3)/ 3! =
280If A,B,C,D,E,F ,G,H,I are the 9 people.
For example, group (ABC) ,(DEF) (GHI) will be counted 6 times like this below.
(ABC) ,(DEF) (GHI)
(ABC),(GHI),(DEF)
(DEF)(ABC)(GHI)
(DEF)(GHI)(ABC)
(GHI)(ABC)(DEF)
(GHI)(DEF)(ABC)
To avoid this repeated counting, we are dividing by 3! i.e 6.
Option A is the right answer.
I hope this will help .
Thanks,
Clifin J Francis,
GMAT SME
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