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13 Sep 2019, 12:58
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hemanthp
Asked: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?
Number of ways = (9C3 * 6C3 * 3C3)/3! = (84 * 20 * 1)/6 = 1680/6 = 280
IMO A
27 Aug 2021, 23:31
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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! = 280
If 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
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! = 280
If 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
17 Mar 2025, 16:11
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KarishmaB
Hi Karishma,
I am still not able to understand why are we specifically dividing by 3! only. We have used the selection formula above, how are the groups getting ordered then? Can you please explain
