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July 4th, 2010, 07:10 AM   #1
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Clarification for a 'counting' problem.

1. In how many ways can you wear 4 different rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand? You can wear as many rings as you want you want on a finger.

2. In how many ways can you wear 4 identical rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand? You can wear as many rings as you want you want on a finger.

Please check if my solution is correct.

First divide for into sum of positive integers. The possibilities are:

4
3,1
2,2
1,1,2
1,1,1,1

Number of ways of choosing one finger for placing all rings= 5C1 = 5
Number of ways of choosing two fingers for placing rings in '3,1 way'= 5C2 = 10
Number of ways of choosing two fingers for placing rings in '2,2 way'= 5C2 = 10
Number of ways of choosing three fingers for placing rings in '1,1,2 way '= 5C3 = 10
Number of ways of choosing four fingers for placing rings in '1,1,1,1, way'= 5C4 = 5

Once the fingers for placing rings have been chosen, determine the order of number of rings on a particular finger

For 4 rings on one finger, number of ways = 1!
For "1,3 way" on two fingers, number of ways = 2!
For "2,2 way" on two fingers, number of ways = 2!/2! = 1
For "1,1,2 way" on three fingers, number of ways = 3!/(1!2!) = 3
For "1,1,1,1 way" on four fingers, number of ways = (4!/4!) = 1

*We will stop our calculations here if the rings are identical*

Now we have chosen particular combination (ex "2 +2 +1"), also we have chosen the fingers to put them on, and also the order (ex: which two fingers will have 2 rings and which one will have 1). Once having done that, you have have to arrange the rings. Which will be in 4! ways for each and every combination.

Answer for problem 1 (identical rings) = 1*5 + 2*10 + 1*10 + 3*10 + 1*5 = 70
Answer for problem 2 (all rings are different) = 70*4! = 1680

Is this solution correct. I know the solution of second part is correct. Because answer can be found using a formula for distributing 'n' identical things amongst 'r' people, every person can get any number of things.

The formula is (n+r-1)C(r-1)
Here, n=4, r=5
8C4 =70.

Query : How has this general formula been derived?
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July 4th, 2010, 07:38 AM   #2
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Re: Clarification for a 'counting' problem.

Pick a ring.
How many fingers can then be chosen to put this ring on?
Do this four times.
5*5*5*5.
Unless I'm missing something (like having all the rings on your thumb is seen as equivalent to having them all on your index finger...)
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July 4th, 2010, 09:23 PM   #3
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Re: Clarification for a 'counting' problem.

@swordfish: your calculations are correct.

One explanation for the formula is that there is a one-to-one correspondence between ways of distributing identical objects among distinguishable containers and binary strings representing those distributions. Suppose 15 identical objects are given to 8 different people, e.g.: 3,4,0,2,1,0,0,5

Notice that there are 7 commas. Replace the figures with tallies: 111,1111,,11,1,,,11111 and it should be clear that every permutation of this string (15 ones and 7 commas) corresponds to a valid distribution of objects.

There are 22 characters, of which 15 are the same and 7 are the same, so there are permutations. Start this line of reasoning again with n objects and r people and you will get the formula you mention.

@chaz: that doesn't work because most arrangements require rings to be put on in a different order.
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July 5th, 2010, 04:20 AM   #4
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Re: Clarification for a 'counting' problem.

I was gonna say... it seemed like much ado about nothing!
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July 5th, 2010, 04:15 PM   #5
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Quote:
Originally Posted by aswoods
@chaz: that doesn't work because most arrangements require rings to be put on in a different order.
That's a bit unclear. If ring 1 is put on finger 1, ring 2 on finger 2, ring 3 on finger 3 and ring 4 on finger 4, there is only one arrangement, but 4! ways of achieving it (assuming some simple rules of placement).
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July 5th, 2010, 05:41 PM   #6
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Re: Clarification for a 'counting' problem.

Permutation/combination problems (as a whole) always bug me with their ambiguity!
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