Find the number of distinct throws which can be thrown with $n$ six faced normal dice which are indistinguishable among themselves. Find the number of distinct throws which can be thrown with $n$ six faced normal dice which are indistinguishable among themselves.
The total outcomes will be $6^n$. But this this has many cases repeated since the dice are indistinguishable among themselves.
 A: Consider six faces as six beggars and n identical dice to be identical coins.
Now,number of distributions is C(n+6-1,6-1)=C(n+5,5).
If a beggar(say face 6)gets no coin,then it is equivalent to anything which does not appear on the dice.
A: Here we can come up with an equation that maps one-to-one with the requirements provided by the problem statement.
Each dice has 6 faces - $1, 2, 3, 4, 5$ and $6$. Now, let us map each of these numbers to alphabets. We can represent $1, 2, 3, 4, 5$ and $6$ by $a, b, c, d, e$ and $f$ respectively.
Now, we can observe that when $n$ dices are thrown, the following maybe the possible outcomes:
$$(1,1,1,1,1...n \ \text{times}) \ \text{or} \ (1,2,2,2,2,...n \ \text{times}) \ \text{or maybe} \ (5,5,3,3,3....n \ \text{times})$$
So whatever the outcome is, each of the digits have to turn up anywhere between $0$ to $n$ number of times. For instance, in the second example, the digit $1$ turns up once and the digit $2$ turns up $(n-1)$ times and all the other digits turn up zero times. Notice that the sum of occurrences of all the digits is always $n$.
We can transform this information into an equation.
$$a + b + c + d + e + f=n$$
where $a$ represents the number of times $1$ occurs in an outcome, $b$ represents the number of times $2$ occurs in an outcome and so on and so forth.
The number of possible outcomes(throws) will be the number of solutions to this equation.
Now, if this is a familiar situation for you, you will immediately get to the answer $\binom{n+5}{5}$, following the formula $\binom{n+r-1}{r}$.
But if you are not, continue reading.
Let us represent the number of occurrence of the digits using binary digits (not binary numbers). For example, in my third example, $5$ occurs twice, so we can represent it by two $1$'s: $1 \ 1$. Now realise that if we represent all of the six digits this way, we will need a total of $n$ number of $1$'s to satisfy the equation. We can separate the digits using $0$'s.
Consider the following example for clarity:
$$\underbrace{\underbrace{11111}_{a = 5} \ \underbrace{0}_\text{separator} \ \underbrace{1111}_{b=4} \ 0 \ \underbrace{1}_{c=1} \ 0 \ \underbrace{11}_{d=2} \ 0 \ \underbrace{}_{e = 0} \ 0 \ \underbrace{1111...(n - 12) \ \text{times}}_{f=n-12}}_{a+b+c+d+e+f=n}$$
Notice that all of the possible solutions are just the different arrangements of $n$ number of $1$'s and $5$ number of $0$'s(why 5? count the number of '$+$' signs in the equation). Therefore, using the formula for arrangement, we can say,
$$\text{Total possible number of arrangements} = \frac{(n+5)!}{n!5!}$$
as there are a total of $(n+5)$ stuff(number of $1$'s and $0$'s), and $n$ alike stuff(the $1$'s) and $5$ alike stuff(the $0$'s).
Now, $$\frac{(n+5)!}{n!5!} = \binom{n+5}{5} = \binom{n+5}{n}$$
