Is there a mathematical expression for this?

Question

1, 3, 10, 35, 126, 464, 1716.

So this is a pattern. All distinct combinations possible with one one-sided die (1), All the combinations possible with two two-sided die (11, 12, 22), All the combinations possible with three three-sided die (111, 112, 113, 122, 123, 133, 222, 223, 233, 333), and so on.

Oh yeah, does anybody venture to find out what it is for eight, eight sided die!

Answer 1

There is. And you can leave the dice out of it. That sequence is simply the number of combinations for one digit, two digits, three digits, etc.

The formula to use is one for combinations with repetition.
(n+r-1)! / (r! * (n-1)!)

In this case, the number of elements you're choosing from (n) is the same as the number of elements you're choosing (r), so the formula can be simplified/reduced to:
(2n-1)! / (n! * (n-1)!)

For your question on eight, it works out as follows:
15! / (8! * 7!)
(15*14*13*12*11*10*9) / (7*6*5*4*3*2)
(13*12*11*10*9) / (6*4)
6435 is your answer.

Answer 2

Here's an answer with actual explaination, not just a random formula pulled out of thin air. You know, in case that matters to you. Since what mister smarty pants above did was look up the work of someone else, who has solved this problem using the method below.

What you want is possible die rolls if order does not matter. There are n^n if we assume the order matters. For example:

11, 12, 21, 22

That's 2² = 4. But there's a repeat in there, since 12 is the same as 21. So what you want is only 3.

So what we have to do is figure out how to get rid of repeats. In order to do that, we have to establish what is repeated. We will use a method known as group actions. This is a very sophisticated method, but this is not a simple problem.

We define two "actions" on an ordered roll. These actions will preserve the outcomes - and thus the corresponding unordered roll will be the same. If we can figure out how to count up all of the equivalent ordered rolls, we know how many unordered rolls there are (since each unordered roll represents a set of equivalent ordered rolls).

T will be a "swap" of the first and second dice.

T: 213 &rarr: 123

R will be a "rotation" of the dice to the left:

R: 51335 → 13355

These two operations generate a group:
http://en.wikipedia.org/wiki/Group_%28mathematics%29

What we want to count is the number of "orbits" or "classes" of equivalent ordered rolls - each of these corresponds to a particular "orderless" roll. So if we count the orbits, we're done.

So how do we do that?

We have to take all of the possible actions (T, R, and combinations of T and R) and figure out which kinds of rolls they do not change. For example:

T: 112 → 112

R: 111 → 111

If we add up the total number of unchanged ordered rolls, for each possible combination of T and R, we're halfway there.

The other half of the problem is finding the size of the group generated by T and R. See:
http://en.wikipedia.org/wiki/Generating_set_of_a_group
http://en.wikipedia.org/wiki/Burnside%27s_lemma

What's great is that this is the ENTIRE symmetric group (you get every possible mix up of the original ordered roll). See:
http://en.wikipedia.org/wiki/Generating_set_of_a_group#Examples

So the size of the group is n!

Now we're back to the first part: counting all the rolls fixed by all the different actions. So how many ordered die-rolls are fixed by any particular element of this group (any particular action)?

I've already gone into this pretty in depth, so I'm going to gloss over the next part. I won't explain in depth, but there are (2n+1)! / (n+1)! total.

So we know that the answer you want is:

(2n+1)! / ( n! (n+1)! )

Which gives:

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220

The 8th term is 6435.

Note: You have a computational error for the 6th term. It is not 464, but 462.

Answer 3

I am in awe of Head Scratcher's mathematical prowess. Nice work.

I couldn't figure it out, but I did use his/her formula to generate a bunch of terms in the series, discovering a typo in your original sequence (464 should be 462).

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410