Probability problem that includes solutions to the linear equation. Let's say that we have a set:
$\Omega=\big\{(x_1,x_2,...,x_{10}):x_1+...+x_{10}=20,x_i \in\mathbb{N} \cup\small\{0\small\}, i \in1,..10\big\}$
One element of this set $(x_1, x_2,...,x_n)$ is chosen randomly, find the probability that $x_1\geq3$ and $x_{10} \geq 5$ 
Now, this is somehow a non-standard problem for me. Anyway, I found a hint in my workbook that U should firstly find the numbers of elements for this set by looking at its generalization $$x_1+...+x_k=n,$$  and then when I find the number of elements of this set, I could easily set $n=20, k=10$ and that would tell me the number of elements for this set.  It says that number of elements of such set is equal to number of ways that we can place $n$ zeros between $k-1$ numbers one (I absolutely have no clue what this means).
Then it says that this is equal to number of ways that out of $n+k-1$ elements we choose $k-1$ of them $n+k-1\choose k-1$. I understand the last part itself, but I have no idea how to get there, and that's just the first half of the problem (without restrictions). Any explanations appreciated.
 A: For the intuition, imagine $n$ "$*$"-s (stars) in a row and you have $k-1$ "$|$"-s (bars). Place the bars between the stars, eg $$**|***|*.$$ The example I've just given shows one way to partition $\{1,...,6\}$ into $3$, namely $$\{1,2\} \cup \{3,4,5\} \cup \{6\}.$$
You then take your $x_i$-s to be the sizes of these partitions, ie above $x_1 = 2$, $x_2 = 3$ and $x_3 = 1$ giving $x_1 + x_2 + x_3 = 6$. (Note that the partitions need not be non-empty: this corresponds to $x_i = 0$.)
We can now make this into a proof. We understand (hopefully!) that partitioning is equivalent to placing $k-1$ bars between $n$ starts. This means that we have a total of $n+k-1$ objects, and we wish to specify the position of $k-1$ of them (ie the bars); the remaining $n$ objects (ie the stars) can be placed anywhere in the remaining places. But fixing $k-1$ of $n+k-1$ objects is precisely what the binomial coefficient does: so the number of different ways is indeed $$\binom{n+k-1}{k-1}.$$
The above links show you how to calculate how many different ways there are of placing the bars. (Note that placing $k-1$ bars gives a partition into $k$ sets, just like a fence with $k-1$ bars in it has $k$ poles.)

There is still work to do: we've only found the size of the set; we now need the size of the subset of interest, and then look at the ratio to get the probability.
When we insist that $x_1 \ge 3$, consider writing $x_1 = x_1' + 3$ with $x_1' \ge 0$; similarly write $x_{10} = x_{10}' + 5$ with $x_{10}' \ge 0$. Set $x_i' = x_i$ for all $i \in \{2,...,9\}$. Then
$$
x_1 + \cdots + x_{10} = 20
\iff
x_1' + \cdots + x_{10}' = 12.
$$
So we've reduced the restricted case back to the general case, but with $20$ replaced by $12$.

Thus we have found our probability: it is
$$
\binom{12+10-1}{10-1} \Big/ \binom{20+10-1}{10-1}.
$$
(You can simplify this further with some algebra, but I don't have a pencil and paper at hand, so leave that part to you!)
I leave generalisations of this to you, should you so desire.

I hope this is helpful for your learning! :)
