Certain number of people arranged in several groups

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Homework Help Overview

The problem involves determining the number of ways to divide 9 people into two groups: one group of 6 people and another group of 3 people. The original poster presents a calculation based on arrangements but arrives at a different answer than expected.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of selecting groups, with some suggesting that the problem is fundamentally about choosing a subset of people. Others question the original poster's approach to counting arrangements and suggest reconsidering how groups are formed.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the nature of group selection, but there is no explicit consensus on the correct method or outcome.

Contextual Notes

There appears to be confusion regarding the treatment of arrangements and the implications of grouping people versus simply selecting them. The original poster's calculation method is under scrutiny, and assumptions about group identity and arrangement are being questioned.

desmond iking
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Homework Statement


Find the number of ways of 9 people can be divided into two groups of 6 people and 3 people.


Homework Equations





The Attempt at a Solution



my working is there are 6! arrangement for 6 people and 3! arrangment for 3 people. then the group of 3 people can be placed at right or left only . so that the 6 people form another group will not be separated. so my working = 3! x 6! x2 = 8640..

the ans given is only 84
 
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This is only a matter of choosing three people that you tell to step away from the others.
 
Fredrik said:
This is only a matter of choosing three people that you tell to step away from the others.

can you explain further?
 
desmond iking said:
can you explain further?
I can't say much more without completely solving the problem for you. Would you agree that if you move 3 people away from the others, you have divided the original group into two, one with 3 people and one with 6? This is of course not the only way to do it. You don't have to separate them physically. You can e.g. hand out funny hats to 3 of them. But no matter what you do, it's a matter of choosing 3 people (or 6 people).
 
No i mean I move the 3 people in a group as 1 item in the arrangement. And the same thing for the 6 people.
 
I don't quite understand what you mean by that. The fact that your calculation includes 6! and 3! suggests that you're taking into account the number of ways that each group can be rearranged, but you're not doing it correctly. How many ways are there to divide 3 people into two groups of 1 and 2? Your method yields 2!·1!·2=4, but the right answer is clearly 3. How many ways are there to divide 4 people into two groups of 2 each? Your method yields 2!·2!·2=8, but the correct answer is 6. That's the number of ways you can distribute 2 identical funny hats among 4 people.
 

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