Permutation, Combination of AAB

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Permutation, Combination of AAB

by pagalmes » Tue Jan 19, 2010 12:58 pm
Hello,

Below are written a few questions that can be solved simply by couting, but I was wondering how to solve them with the classic formulas:

1 - How many ordered arrangements of 2 letters picked from AAB can be done?
2 - How many unordered arrangements of 2 letters picked from of AAB can be done?
3 - How many unordered arrangements of 4 letters picked from of AAABBC can be done?

Answers ?

1 - The answer seems to me to be: {AA, AB, BA}. So that's 3
The permutation formula gives:

3!
________ = 3
2! (3-2)!

2! is because you have twice the letter "A". The result is the same.

2 - The answer seems to me to be: {AA, AB}. So that's 2.
The combination formula gives:

3!
________ = 3 (which is different from 2!)
2! (3-2)!

Where is my mistake? How should I write the problem ?

3 - Any idea?

Thanks[/b]
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by harsh.champ » Wed Feb 03, 2010 4:55 am
pagalmes wrote:Hello,

Below are written a few questions that can be solved simply by couting, but I was wondering how to solve them with the classic formulas:

1 - How many ordered arrangements of 2 letters picked from AAB can be done?
2 - How many unordered arrangements of 2 letters picked from of AAB can be done?
3 - How many unordered arrangements of 4 letters picked from of AAABBC can be done?

Answers ?

2 - The answer seems to me to be: {AA, AB}. So that's 2.
The combination formula gives:

3!
________ = 3 (which is different from 2!)
2! (3-2)!

Where is my mistake? How should I write the problem ?

[/b]
______________
Hi pagalmes,
In your 2nd answer both the A's are not differentiable,hence you are taking A(1)B and A(2)B as different entities.
In these kind of questions the best way is to consider identical and different entities separately.
suppose 5B and 2A are there then you have 3 arrangements.
Last edited by harsh.champ on Wed Feb 03, 2010 5:07 am, edited 2 times in total.

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by harsh.champ » Wed Feb 03, 2010 5:04 am
pagalmes wrote:Hello,

Below are written a few questions that can be solved simply by couting, but I was wondering how to solve them with the classic formulas:

1 - How many ordered arrangements of 2 letters picked from AAB can be done?
2 - How many unordered arrangements of 2 letters picked from of AAB can be done?
3 - How many unordered arrangements of 4 letters picked from of AAABBC can be done?

Answers ?

3 - Any idea?

Thanks[/b]
______________
Now over here,we have to consider cases:-
Case 1:-3As
Hence,the 4th letter can be B or C(2 counts)
Case 2:-2As
so,rest of the selections can be (B,B);(B,C)
Case 3:-1As
here we have only 1 selection (BBC)
The case of 0As does not exist.
Hence,adding all the above 3 cases (2+2+1),we get the answer as 5.