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]
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]
