What does bijection mean with regards to Combinatorics and Sets? I think I understand the 2 conditions that are necessary for a function to be bijective, but in a book of Combinatorics I am reading it talks about Bijections with sets and with combinatorial problems, and doesn't explain what it is very well in my opinion. It simply says it's a one-to-one correspondence between different items... What does a bijection mean then with regards to Combinatorics and sets?
For example, earlier in the book it gave this problem:

Suppose that $a,~b,~c,~d$ and $e$ are positive integers. How many solutions are there to the equation $$a+b+c+d+e=11$$?

This can be found by considering $11$ items and $4$ gaps between different items out of the $10$ possible gaps, so the solution is $\binom{10}{4}$. I understand this perfectly.
Later, when talking briefly about bijections it says of the aforementioned problem that the coding (of $4$ objects that can be chosen of one and $6$ of the same type of object that can't be chosen (in this case, gaps)) is 'unique and reversible, or, in other words, that it represented a bijection.'
What does it mean by this? What bijection is it talking about? I have no idea. Also, I don't understand what it means by reversible.
Thank you for your help in this very basic question.
 A: Let me see if I can show you in detail what’s going on behind the scenes, so to speak, in that problem and the book’s explanation.
Let
$$S=\{\langle x_1,x_2,x_3,x_4,x_5\rangle\in\Bbb Z^+:x_1+x_2+x_3+x_4+x_5=11\}\,;$$
we want to know $|S|$. The idea behind the solution is to find a set $A$ whose cardinality is easier to determine and show that $|A|=|S|$ by showing that there is a bijection between $A$ and $S$.
In this case we imagine lining up $11$ items: $c_1,c_2,c_3,c_4,c_5,c_6,c_7,c_8,c_9,c_{10},c_{11}$. We let $G$ be the set of gaps between adjacent items; clearly $|G|=10$. Finally, we let $A=\{X\subseteq G:|X|=4\}$, the set of $4$-element subsets of $G$; we know that $|A|=\binom{10}4$. If we can find a bijection between $A$ and $S$, we’ll have shown that $|S|=\binom{10}4$.
And you already know what the bijection is: if $s=\langle x_1,x_2,x_3,x_4,x_5\rangle\in S$, we let $f(s)=\{g_1,g_2,g_3,g_4\}\in A$, where $g_1$ is the gap between $c_{x_1}$ and $c_{x_1+1}$, $g_2$ is the gap between $c_{x_1+x_2}$ and $c_{x_1+x_2+1}$, $g_3$ is the gap between $c_{x_1+x_2+x_3}$ and $c_{x_1+x_2+x_3+1}$, and $g_4$ is the gap between $c_{x_1+x_2+x_3+x_4}$ and $c_{x_1+x_2+x_3+x_4+1}$, so that there are $x_1$ items before the gap $g_1$, $x_2$ items between gaps $g_1$ and $g_2$, $x_3$ items between gaps $g_2$ and $g_3$, $x_4$ items between $g_3$ and $g_4$, and $x_5$ items after gap $g_4$. Clearly this set of gaps is completely determined by the solution $s$: given $s$, there is a unique set of $4$ gaps described in this way by $s$. This simply means that $f$ is a function from $S$ to $A$ and is what the book’s unique is getting at.
Reversible simply means that the function $f$ has an inverse, i.e., that it is a bijection: it is a surjection, because every set of $4$ gaps is $f(s)$ for some solution $s\in S$, and it is an injection, because if we are given a set of $X=\{g_1,g_2,g_3,g_4\}$ of $4$ gaps, we can determine the unique $s\in S$ such that $f(s)=X$.
A: It is like speed-dating.
If $10$ women want to date $10$ men simultaneously, you have to match every woman with a man, every man with a woman, such that every man has a woman to date and every woman has a man to date.
To each element of the first set (the women) is associated one and only one element of the second set (the men), and reciprocally.
