How Many Unique Outcomes with 5 Dice Ignoring Order?

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The discussion focuses on calculating the number of unique outcomes when rolling 5 dice, where order does not matter and repetitions are allowed. It highlights that this scenario involves combinations rather than permutations, specifically allowing for repeated elements. The method described uses a "stars and bars" approach, where 5 dice rolls are represented as stars and the boundaries between different numbers as bars. The total arrangements can be calculated as C(10,5), which represents choosing 5 positions for the bars among 10 total spots. This combinatorial method effectively simplifies the problem of counting unique outcomes.
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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