$5$ "identical" dice?

Question

Question in video:

$5$ identical dice are thrown simultaneously. How many different outcomes are possible?

Solution in video:

Here, we are not given different dice, but same dice. So, the outcomes from the dice can't be rearranged.
Then the video proceeds with cases from: $0,1,2\dots5$ different outcomes and solved it.

My question:

In my understanding, if we can read the number of the dice then we can definitely identify/distinguish them and rearrange them. Though when same number appears on more than $1$ dice then the dice with same numbers can't be distinguished. So, I can't really digest the solution given by the video.

Basically what is the difference between different dice and identical dice (in real world?)?
Please help.

Answer 1

With identical dice we are counting combinations - the order of the list of outcomes for each dice does not matter because we cannot assign an outcome to a specific dice. With different dice we are counting permutations - the order of the list of outcomes does matter, because we can assign each outcome to a specific dice.

For example, there is only 1 way to throw 1,2,3,4,5 with five identical dice, but if the dice are different (each has a different colour, for example) then there are 120 ways to throw 1,2,3,4,5.

Answer 2

Addendum added to respond to the comment/question of InanimateBeing.

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Because the $(5)$ dice are presumed identical, the problem becomes more rather than less complicated.

You have to consider how the dice rolls may be grouped, and provide a count, for each grouping. Case work shown below. For each Case, I will use $T_k$ to denote the enumeration for Case $k$.

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$\underline{\text{Case 1: all 5 dice the same number}}$
There are $\displaystyle ~\binom{6}{1}~$ ways of determining which element from $\{1,2,3,4,5,6\}$ will be shown. So,

$$T_1 = 6.$$

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$\underline{\text{Case 2: 4 of one number, 1 of another number}}$
There are $\displaystyle (6 \times 5)$ ways of choosing the number that will show $(4)$ times and then choosing the number that will show $(1)$ time.

$$T_2 = 30.$$

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$\underline{\text{Case 3: 3 of one number, 2 of another number}}$
There are $\displaystyle (6 \times 5)$ ways of choosing the number that will show $(3)$ times and then choosing the number that will show $(2)$ times.

$$T_3 = 30.$$

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$\underline{\text{Case 4: 3 of one number, 1 of another number, 1 of a third number}}$
There are $\displaystyle \left[6 \times \binom{5}{2} \right]$ ways of choosing the number that will show $(3)$ times and then choosing the two numbers that will each show $(1)$ times.

$$T_4 = 60.$$

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$\underline{\text{Case 5: 2 of one number, 2 of another number, 1 of a third number}}$
There are $\displaystyle \left[\binom{6}{2} \times 4 \right]$ ways of choosing the two numbers that will each show $(2)$ times and then choosing the number that will show $(1)$ time.

$$T_5 = 60.$$

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$\underline{\text{Case 6: 2 of one number, 1 of another number, 1 of a third number, 1 of a 4th number}}$
There are $\displaystyle \left[6 \times \binom{5}{3} \right]$ ways of choosing the number that will show $(2)$ times and then choosing the three numbers that will each show $(1)$ time.

$$T_6 = 60.$$

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$\underline{\text{Case 7: 5 different numbers show}}$
There are $\displaystyle (6)$ ways of choosing the number that does not show.

$$T_7 = 6.$$

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$\underline{\text{Final Computation}}$

$$T_1 + \cdots + T_7 = 6 + 30 + 30 + 60 + 60 + 60 + 6 = 252.$$

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Addendum
Responding to the comment/question of InanimateBeing.

... my question has more to do with the dice than the combinations.

Suppose that the $5$ dice are all different colors, so that (for example) having the $(5)$ different numbers of 1-2-3-4-5 can occur in $(5!)$ ways.

Then, the number of possibilities for each of the $(5)$ dice is totally independent of the possibilities for any of the other dice. This is a consequence of the dice being different colors.

In this situation, the enumeration would therefore be

$$(6)^5.$$

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Addendum-2
I thought that it was somewhat coincidental, that the enumeration turned out to be $~\displaystyle \binom{10}{5}.~$ See my comment, following the answer of Robert Shore, which uses Stars and Bars analysis.

For pertinent references, see

• this article

• and this article.

Answer 3

Let $x_i$ be the number of dice that land on $i$. Then you want to know how many solutions there are in non-negative integers to $x_1+x_2+x_3+x_4+x_5+x_6=5$. This is a classic stars-and-bars problem, and the answer is $\binom{10}{5}=252$.

The point of considering the dice to be identical (instead of, say, different colors) is that the only thing that distinguishes two throws is how many different dice give you each result, not which specific dice give you that result.