How Many Unique Ways Can 4 Dice Be Combined?

[nille40]

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

 

[nille40]

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

 

[tacman]

Hi Nille, thank you for sharing this combinatorial problem with us! It's great that you're trying to figure out how to approach it. Let me try to explain the correct solution in a different way that may help you understand it better.

First, let's break down the problem into smaller parts. We have 4 dices, and each dice has 6 possible outcomes (numbers 1-6). So, in total, we have 6 x 6 x 6 x 6 = 1296 possible combinations if the order matters. But since the order doesn't matter, we need to divide this number by the number of ways we can arrange the 4 dices, which is 4! (4 x 3 x 2 x 1).

This gives us 1296 / 4! = 1296 / 24 = 54 unique ways to combine 4 dices. This is the same answer you got, but the approach to getting it was not quite correct.

Now, let's look at the solution you mentioned: {6+4-1} \choose {4}. This is known as the combination formula, and it is used when we want to select a certain number of objects from a larger set, without considering the order. In this case, we have 6 options for each dice, and we want to select 4 dices, so we use the formula {n+r-1} \choose {r}, where n is the number of options and r is the number we want to select.

So, in our case, we have {6+4-1} \choose {4} = {9} \choose {4} = 126. This may seem like a larger number than 54, but remember that the combination formula considers the order of the objects to be irrelevant. In other words, it's also counting combinations like 2-2-2-2 or 3-3-3-3, which are essentially the same as 1-1-1-1 or 6-6-6-6. That's why the answer is different.

I hope this helps clarify things for you. Keep practicing and don't be afraid to ask for guidance when needed. Combinatorics can be tricky, but with practice and patience, you'll get the hang of it. Good luck!