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permutation

by klaud » Sat Feb 04, 2012 12:13 pm
How many ways are there to seat 11 people in a row of 3 chairs?

About this question, the right approach is this: for the first chair i can choose 11 people,for the second 10 and for the third 9.

11x10x9= 990 combinations

Why in the question below i can't use the same approach?

In how many ways can 5 different rings be worn on the any four fingers of the right hand?
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by pemdas » Sat Feb 04, 2012 3:44 pm
klaud wrote:How many ways are there to seat 11 people in a row of 3 chairs?

About this question, the right approach is this: for the first chair i can choose 11 people,for the second 10 and for the third 9.

11x10x9= 990 combinations

Why in the question below i can't use the same approach?

In how many ways can 5 different rings be worn on the any four fingers of the right hand?
first of all these are two different problems with quite different situations. There is no constraint on any fingers wearing rings, so we can have
- either 1 finger with all five rings and 3 fingers bare
- 2 fingers with five rings and so on and on ...

we can choose 4 fingers of 5 ones of the right hand with assumption of the right hand having 5 fingers, choose 2 fingers of 5, etc.

rings are different and fingers are different ~discard tricky notion 'on any four fingers' order matters as fingers are different.
5C4*(4P1*5P1 + 4P2*5P2 + 4P3*5P3 + 4P4*5P4) = 22,900 ways
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by klaud » Sun Feb 05, 2012 12:36 am
pemdas wrote:
klaud wrote:How many ways are there to seat 11 people in a row of 3 chairs?

About this question, the right approach is this: for the first chair i can choose 11 people,for the second 10 and for the third 9.

11x10x9= 990 combinations

Why in the question below i can't use the same approach?

In how many ways can 5 different rings be worn on the any four fingers of the right hand?
first of all these are two different problems with quite different situations. There is no constraint on any fingers wearing rings, so we can have
- either 1 finger with all five rings and 3 fingers bare
- 2 fingers with five rings and so on and on ...

we can choose 4 fingers of 5 ones of the right hand with assumption of the right hand having 5 fingers, choose 2 fingers of 5, etc.

rings are different and fingers are different ~discard tricky notion 'on any four fingers' order matters as fingers are different.
5C4*(4P1*5P1 + 4P2*5P2 + 4P3*5P3 + 4P4*5P4) = 22,900 ways

I found this explanation:

5 rings, 4 fingers
1st ring can be worn on any of the 4 fingers => 4 possibilities
2nd ring can be worn on any of the 4 fingers => 4 possibilities
3rd ring can be worn on any of the 4 fingers => 4 possibilities
4th ring can be worn on any of the 4 fingers => 4 possibilities
5th ring can be worn on any of the 4 fingers => 4 possibilities

Total possibilities = 4*4*4*4*4 = 4^5.
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by pemdas » Sun Feb 05, 2012 8:09 am
@klaud this is wrong explanation, because the rings are different. Five different rings can be worn on one single finger in 5! ways. The explanation you found doesn't address the question and it cannot be correct.
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