How many possible outcomes are there if n identical dice are thrown?

Question

Two outcomes are the same if they contain the same # of 1's, same # of 2's, same # of 3's, same # of 4's, same # of 5's, and same # of 6's

The answer is 21 for n=2 only.

Please note before you answer that I am not asking for the permutations of n dice, or 6^n. As The_Dark_Knight correctly
noticed (while his answer is incorrect), 6^n counts 112222 and 212122 twice, prohibited by the question.

f(3) = 56

Answer 1

The total number of ways the dice could fall is 6ⁿ.

However you say you can't distinguish one pattern from another.. i.e. if you had 6 dice then 112222 is no different from 212122.
That is to say the internal arrangement of the dice doesn't matter and one internal arrangement is indistinguishable from the another.
Therefore the actual number of different outcomes should be
6ⁿ/n!.

I think...

PS: I know my answer is wrong.. and I am not able to figure out why.. grrrr..

PPS: Ok I remember reading this in Feynman's lectures, this is similar to Bose-Einstein distribution, in which case the answer, which I have forgotten how to derive is
(n+g - 1)!/[n!.(g-1)!]

n is the number of particles (dice)
g is the number of energy states (well.. 6 here, the different faces of the die).

So in this particular case the answer is (n+5)!/(n!.5!)

I had to go and read it again.

Answer 2

You are dealing with combinations with repetition allowed.
If n is the number of dice then the number of distinct combinations (outcomes) is given by:

f(n) = C(6+n-1,n) = C(5+n,n) = (5+n)!/n!5!

If n =1, f(1) = 6!/5! = 6 outcomes

If n = 2, f(2) = 7!/2!5! = 7x6/2 = 21

if n= 3, f(3) = 8!/3!5! = 8x7 =56

For fun:

if n= 100 then f(n) = (105)!/100!5! = 96,560,646

Answer 3

for n=1 it would be 6. for n=2 it would be 36. for n=3 it would be 6 times 6 times 6 which equals 216. so it would be 6 to the n. n being the number of dice

Answer 4

you multiply the numbers of the dice by the numbers on the other dice.

So 6 X 6 = 36

Answer 5

The dice would end up in 6^n different outcomes

Answer 6

I think it 30... I forget how to explain but it is

6! over something but I think you only multiply the first two numbers since there are only two dice. or you could figure it by trial and error each number represents one of the dice
11, 12, 13 , 14, 15, 16
21,22, 23, 24, 25,26
31, 32,33, 34, 35, 36
41, 42, 43,44,45,46,
51,52,53,54,55,56
61, 62,63,64,65,66
well what do ya know... my first theory didnt work.. there are thirty-six possibilities through trial and error
or not

Answer 7

OK, I have been working on this for the past 5 min and I came up with this calculation.....

21, I think.

Answer 8

s=sides on the die. n=number dice...so

it would be s!n. the "!" being the symbol for factoral.

Answer 9

more details