If $10$ identical dice are tossed, then how many possible outcomes are there?
The answer to this question is solved by considering it similar to the problem of non-negative integral solutions i.e.,
$$x_1+x_2+...x_n=r.$$
I'm not able to understand how this analogy is made. Can someone please explain this part?
Before we get to your question, let's look at the case of two indistinguishable dice.
Since the dice are identical, outcomes are distinguished by how many times each number appears on the dice.
Suppose both dice are white and the same size, so that they are indistinguishable.
Method 1: There are $\binom{6}{1}$ outcomes in which the same number appears on both dice and $\binom{6}{2}$ outcomes in which different numbers appear on the two dice. Hence, the number of outcomes when the two dice are indistinguishable is $$\binom{6}{1} + \binom{6}{2} = 6 + 15 = 21$$ As a check, note that if we were to paint one of the two white dice red and the other blue, that would not change the outcome of the six cases in which the same number appears on both dice. On the other hand, if the outcomes on the two dice are different, there are two ways to choose which color to paint the die with the larger number. Note that $$\binom{6}{1} + 2\binom{6}{2} = 6 + 2 \cdot 15 = 6 + 30 = 36 = 6^2$$ which is the number of outcomes for two distinguishable dice.
Method 2: Let $x_i$, $1 \leq i \leq 6$, be the number of dice on which the number $i$ appears. Since there are two dice, $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 2 \tag{1}$$ Equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of five addition signs in a row of two ones. For instance, $$1 1 + + + + +$$ corresponds to the solution $x_1 = 2$, $x_2 = x_3 = x_4 = x_5 = x_6 = 0$ (both dice show the number 1), while $$+ 1 + + 1 + +$$ corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 0$, $x_4 = 1$, $x_5 = x_6 = 0$ (one die shows a 2 while the other shows a 4). The number of solutions of equation 1 is $$\binom{2 + 5}{5} = \binom{7}{5} = \frac{7!}{5!2!} = \frac{7 \cdot 6 \cdot 5!}{5! \cdot 2 \cdot 1} = 7 \cdot 3 = 21$$ since we must select which five of the seven positions required for two ones and five addition signs will be filled with addition signs. Note that this answer agrees with the one we obtained in method 1.
Note: Since a particular solution of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the nonnegative integers corresponds to the placement of $n - 1$ addition signs in a row of $k$ ones, the equation has $$\binom{k + n - 1}{n - 1}$$ solutions in the nonnegative integers since we must select which $n - 1$ of $k + n - 1$ positions required for $k$ ones and $n - 1$ addition signs will be filled with addition signs.
Method 1 above would be unwieldy to apply here since there are so many cases. On the other hand, we can use Method 2.
Since the dice are indistinguishable, outcomes are distinguished by how many times each number appears on the dice.
Let $x_i$, $1 \leq i \leq 6$, be the number of times the number $i$ appears on the dice. Since there are ten dice, $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 10 \tag{2}$$ which is an equation in the nonnegative integers. The number of solutions of equation 2 is found by setting $k = 10$ and $n = 6$ in the formula stated above.