How Many Unique Outcomes with 5 Dice Ignoring Order?

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    Combinatorics
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Discussion Overview

The discussion revolves around calculating the number of unique outcomes when rolling 5 dice, where the order of the outcomes does not matter. Participants explore combinatorial methods to address this counting problem, considering the implications of allowing repeated values in the outcomes.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on counting unique permutations of 5 dice rolls, noting their lack of recent experience with combinatorics.
  • Another participant explains that permutations where order does not matter are referred to as combinations, introducing the formula for combinations without repetition.
  • A subsequent reply highlights the challenge of the problem due to the allowance of repeated values in the outcomes, contrasting it with standard combinations.
  • One participant proposes a method using stars and bars to visualize the arrangement of dice rolls, suggesting that the problem can be reduced to arranging bars in a set number of spots.
  • A later reply expresses appreciation for the proposed method, indicating a positive reception to the combinatorial approach presented.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final counting method, as the discussion includes various interpretations and approaches to the problem.

Contextual Notes

Participants acknowledge the complexity introduced by allowing repeated values in the outcomes, which affects the application of standard combinatorial formulas.

gravenewworld
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Say you have 5 regular die, how many permutations are possible if permutations such as 1,1,1,1,2 and 1,2,1,1,1 are not unique but considered the same (ordering doesn't matter)? I haven't done any combinatorics work in almost 6 years, so I am completely rusty on counting problems. No this, isn't homework, I graduated 2 years ago, I am just trying to help out a college friend of mine who is a computer programmer.
 
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A permutation where ordering doesn't matter is just a 'combination'

To choose k objects from n, without respect to order, there are

_{n}C_k = \frac{n!}{k!(n-k)!}

ways to do it. (Where 'n C k' is just a symbol for the number of combinations)

See wikipedia for more: http://en.wikipedia.org/wiki/Combination
 
Last edited:
Mute said:
A permutation where ordering doesn't matter is just a 'combination'

To choose k objects from n, without respect to order, there are

_{n}C_k = \frac{n!}{k!(n-k)!}

ways to do it. (Where 'n C k' is just a symbol for the number of combinations)

See wikipedia for more: http://en.wikipedia.org/wiki/Combination

But for a combination, according to wiki, 'the elements of the combination can not be repeated'. This is what makes this problem harder, because you are allowed to have repitition for example 1,1,1,1,1 is one combination as well as 1,1,1,1,2=1,1,1,2,1=1,1,2,1,1=1,2,1,1,1=2,1,1,1,1 etc.



So I have the set {1,2,3,4,5,6} and must choose 5 times from it. Repetition is allowed and ordering doesn't matter.
 
Two sequences are equal if they have the same amount of each number. Consider the rolls
1,2,3,4,5
1,2,3,3,4
6,5,4,3,2
If we let a star represent each die roll and separate each 1,2,3,4,5,6 by a bar.
*|*|*|*|*| represents the first roll
*|*|**|*|| represents the second, and
|*|*|*|*|* represents the third.

There are then 10 total spots to arrange the 5 bars and 5 stars.
_ _ _ _ _ _ _ _ _ _
If we leave the spots where the we place the bars as blank spots, the problem reduces to the number of ways to arrange 5 bars in 10 spots or C(10,5)
 
Last edited:
Vid said:
Two sequences are equal if they have the same amount of each number. Consider the rolls
1,2,3,4,5
1,2,3,3,4
6,5,4,3,2
If we let a star represent each die roll and separate each 1,2,3,4,5,6 by a bar.
*|*|*|*|*| represents the first roll
*|*|**|*|| represents the second, and
|*|*|*|*|* represents the third.

There are then 10 total spots to arrange the 5 bars and 5 stars.
_ _ _ _ _ _ _ _ _ _
If we leave the spots where the we place the bars as blank spots, the problem reduces to the number of ways to arrange 5 bars in 10 spots or C(10,5)

Hey that's a pretty neat way of thinking about it! Thanks.
 

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