How can I solve $1 \leq a < b-2 < c-4 < d-6 < e-8 \leq 22$? I've just learned how to find the number of integer solutions of 
this kind of inequations,
$$x_1 + x_2 + \dots + x_k = n, \qquad(x_i\geq0)$$
which is $\binom{n+k-1}{k-1}$. But I have no idea about solving these ones:
$$ 1 \leq a < b-2 < c-4 < d-6 < e-8 \leq 22. \tag{1}$$
The answer is $\binom{22}{5}$, but how?!
Or solving:
$$ 4x_1 +  3x_2 = n.\quad (x_i\geq0)\tag{2}$$
I think these are classic and well-known inequalities but I can't find anything about them in my Discrete Mathematics books :(
Please somebody teach me about these two forms of questions.
Thanks in advance.
 A: I'll give a sketch on how one could possibly think of it. From the given inequalities, we have $a\geq1, b\geq4, c\geq7, d\geq10, e\geq13$ and $e\leq30$. Furthermore,
$$1\leq a\leq b-3\leq c-6\leq d-9\leq e-12\leq18.$$
Consider the following diagram:
$$1\underset{\smile }{\;\;\;}a\underset{\smile }{\;\;\;}b-3\underset{\smile }{\;\;\;}c-6\underset{\smile }{\;\;\;}d-9\underset{\smile }{\;\;\;}e-12\underset{\smile }{\;\;\;}18.$$
The curved braces below represent the difference between the adjacent numbers. Note that any insertion of nonnegative numbers at these braces uniquely corresponds to a choice of $a,b,c,d$ and $e$. For example, take $a=2,b=6,c=11,d=16,e=23$. We could represent this choice as
$$1\underset{+1}{\;\;\;}a\underset{+1}{\;\;\;}b-3\underset{+2}{\;\;\;}c-6\underset{+2}{\;\;\;}d-9\underset{+4}{\;\;\;}e-12\underset{+7}{\;\;\;}18.$$
So, we reduced counting the number of solutions to the inequality to counting the number of ways to insert numbers between $1,a,b-3,c-6,\ldots$ Let's call these differences $v_1,v_2,\ldots,v_6$. So in general we have
$$1\underset{+v_1}{\;\;\;}a\underset{+v_2}{\;\;\;}b-3\underset{+v_3}{\;\;\;}c-6\underset{+v_4}{\;\;\;}d-9\underset{+v_5}{\;\;\;}e-12\underset{+v_6}{\;\;\;}18.$$
Clearly, we have to count how many ways there are to choose $6$ numbers $v_1,\ldots,v_6$ such that their sum is $17$. As you know, this is exactly
$${17+6-1\choose6-1}={22\choose5}.$$

I'll go one with the equation $4x_1+3x_2=n$. Actually, there isn't a beautiful answer to this one.
An integer $x_1$ will induce an integer solution for $x_2$ if and only if $n-4x_1$ is nonnegative and divisible by $3$. Let's look at a few examples and see what happens:
If $n=3$, then obviously we have $x_1=0$, $x_2=1$. Clearly any mutiple of $3$ will give at least one solution.
Take $n=15$. There are two possibilities: $x_1=0$ or $x_1=3$. It is not very surprising that $x_1$ is a multiple of $3$: if $15-4x_1$ is divisible by $3$, then so is $x_1$.
Similarly, if $n=78=26\cdot3$, then any $x_1$ for which $78-4x_1\geq0$ and $x_1$ is divisible by $3$ will give a solution. So by substracting $4x_1$ we essentially substract a multiple of $12$. Hence, the question becomes how many multiples of $12$ there are in $[0,78]$. This is given by $\left\lfloor\frac{78}{12}\right\rfloor+1=7$, where $\left\lceil a\right\rceil$ denotes the smallest integer greater then or equal to $a$. (If it is not very clear why it is $\lfloor\frac n{12}\rfloor+1$ and not $\lfloor\frac n{12}\rfloor$ or $\lceil\frac n{12}\rceil$, try to find out what the formula should be when $n$ is a multiple of $12$.)
If $n$ is not a multiple of $3$, the method is quite similar. If $n=3x+1$, then we can substract $4$ and obtain $3(x-1)$. Look what happened, we reduced it to the case where $n$ is a multiple of $3$. So the answer here is $\left\lfloor\frac{n-4}{12}\right\rfloor+1$.
If $n=3k+2$, then all we have to do is substract $4$ two times and obtain $\left\lfloor\frac{n-8}{12}\right\rfloor+1$. Of course, there may be some issues when $n-8<0$ or $n-4<0$ in the previous case. Luckily, there aren't ;-) If $n-8<0$ the formula would give $-1+1$ which is $0$. Indeed, there are no solutions in this case. Similarly for $n-4<0$.
Note that all this case-working can be shortly summarized. If $n-4$ or $n-8$ is not a multiple of $12$, then $\lfloor\frac{n-4}{12}\rfloor=\lfloor\frac n{12}\rfloor$ and $\lfloor\frac{n-8}{12}\rfloor=\lfloor\frac n{12}\rfloor$ respectively. This means the answer will just be
$$\left\lfloor\frac n{12}\right\rfloor+1$$
no matter which case we have, as long as $n\geq12$.
A: Let us take the one in the title of your post $$1 < a < b-2 < 10$$ Just look at the inequalities from left to right :  
$a$ is greater than $1$ but smaller then $b-2$
$b-2$ is greater than $1$ so $b$ is greater than $3$
$b-2$ is smaller than $10$ so $b$ is smaller than $12$
so $a$ is smaller than $10$ but greater than $1$  
Does this make things clearer ?
