How Many Unique Ways Can 4 Dice Be Combined?

  • Context: Undergrad 
  • Thread starter Thread starter nille40
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the number of unique combinations of values when rolling 4 dice, where the order of the dice does not matter. The initial incorrect approach suggested using \(6^4\) combinations divided by \(4!\) to account for order, which is not valid. The correct method involves using the combinatorial formula \({{n+k-1} \choose {k}}\), specifically \({{4+6-1} \choose {4}}\), to determine the number of ways to select 4 values from 6 possible outcomes with replacement.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with the concept of combinations and permutations
  • Basic knowledge of probability theory
  • Ability to manipulate binomial coefficients
NEXT STEPS
  • Study combinatorial formulas, particularly the stars and bars theorem
  • Explore the concept of generating functions in combinatorics
  • Learn about probability distributions related to dice rolls
  • Investigate advanced counting techniques in discrete mathematics
USEFUL FOR

This discussion is beneficial for mathematicians, educators, students studying combinatorics, and anyone interested in probability theory and its applications in games involving dice.

nille40
Messages
34
Reaction score
0
Hi all!

In how many unique ways can 4 dices be combined? Note that the order amongst the dices is not relevant, so 1-2-3-4 = 4-3-2-1.

My idea is that you select the values, one by one. You can select the first value in 6 ways, the second in 6 ways, the third in 6 ways and the fourth in 6 ways. This yield [tex]6^4[/tex] combinations. The order was irrelevant, so the answer should then be [tex]\frac{6^4}{4!}[/tex].

This is obviously wrong... I'm trying to figure out how to think to solve a problem like this.

The answer is

[tex]{6+4-1} \choose {4}[/tex]

Which basically means "select 4 of the 6, and put each value back when you've selected it". I don't get this...

Would really appreciate some guidance!
Nille
 
Physics news on Phys.org
Ok, I have an idea.

Lets say we have dices in a line. The first dice has the value 1, the second 2, the third 3...the sixth 6. This yields the equation

[tex]x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4[/tex]

So the solution [tex]x_1 = 2, x_3 = 1, x_4 = 1[/tex] means that two dices has the value 1, one has 3 and one has 4.

This equation has the solution
[tex]{{4 + 6 - 1} \choose {4}} = {{4+6-1} \choose {5}}[/tex]

Can this be solved in some other way?
Nille
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 41 ·
2
Replies
41
Views
7K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K