# Clarification for a' counting' problem.

• Jul 3rd 2010, 07:55 PM
swordfish774
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?
• Jul 4th 2010, 10:30 AM
oldguynewstudent
Quote:

Originally Posted by swordfish774
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?

What you are describing is the distribution of 4 distinct objects to 5 distinct recipients with no restrictions on how many objects a recipient can receive for the first problem. The answer is $5^4$.

For the second problem it is 4 identical objects to 5 distinct recipients. This answer is $\left(\left({5\atop 4}\right)\right) = \left({5 + 4 - 1\atop 4}\right) = 70$
• Jul 4th 2010, 07:46 PM
swordfish774
Quote:

What you are describing is the distribution of 4 distinct objects to 5 distinct recipients with no restrictions on how many objects a recipient can receive for the first problem. The answer is $5^4$.
This is wrong. Consider the case, where you put two distinct rings on one particular finger. You can even arrange those two rings.

Answer to the second part is right. My question is, how do you get the formula? What is the logic and calculation behind it? I don't understand this symbol: $\left(\left({5\atop 4}\right)\right) = 70$

Thanks.
• Jul 5th 2010, 11:46 AM
chaoticmindsnsync
oldmannewstudent is correct. The answer to the first problem (different rings) is $5^4$. Think about it this way. You're creating a 4-list from a set with 5 elements.

In the second problem (identical rings), we are creating multisets. A multiset is like a set except repeated elements are allowed. Thus, the notation $\left(\binom{5}{4}\right)$ means 5 "choose" 4, but we are allowed to choose the same number more than once. Thus, we have $\{1,1,1,1\}$ as an example. While a proof is possible, an example illustrates better.

This link on wikipedia should give you a decent example: Multiset - Wikipedia, the free encyclopedia

Hope that helps.
• Jul 5th 2010, 12:02 PM
swordfish774
@chaoticmindsnsync

Consider, two fingers (thumb, index) and two rings (red, blue). Answer according to above explanation, answer would be 2^2 =4.

But here are the 6 possibilities:
1. thumb : red, index : blue
2. thumb : blue, index : red
3: thumb : blue, red (blue is put in first then the red one), index: none
4. thumb : red, blue (notice it's different from above one), index : none
5. index : red, blue, thumb : none
6: index : blue, red, thumb : none
• Jul 5th 2010, 12:07 PM
swordfish774
@chaoticmindsnsync
As to the other problem, I interpreted your wording different in the question. To me, we were mapping the rings to the fingers, and there are $5^4$ such mappings. To give you an example of what I was thinking, if we let $A=\{1,2,3,4\}$ and $B=\{A,B,C,D,E\}$, then the number of functions $f:A\rightarrow B$ is $5^4$. Then, in your previous post, you mentioned your were considering the condition that evaluating $f(1)=A$ first and $f(2)=A$ second is different that evaluating $f(2)=A$ first and $f(1)=A$ second. While absurd for functions, this a valid question for different rings on a hand. And after some thought, I agree with you. If we consider putting a blue ring on first than a red ring on the same finger is different than putting the red on first then the blue second, then for $n=5,k=4$, there are $70\cdot 4!$ ways to arrange the rings (since we are turning the combination back into a permutation).