<s>[INST] Number of distinct arrangement of $(a,b,c,d,e)$ 
If $a<b<c<d<e $ be


positive  integer such that


$a+b+c+d+e=20$.


Then number of distinct


arrangement of $(a,b,c,d,e)$ is

Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \  1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\ 4\ 5\ 6$
Now after that solution given in book as
Total number of ways
$ \displaystyle =\binom{4}{0}+\frac{\binom{4}{1}}{4}+\frac{\binom{4}{2}}{3}+\frac{\binom{4}{3}}{2}+\frac{\binom{4}{4}}{1}$
$\displaystyle = 1+1+2+2+1=7$
I did not understand last $2$ line
i  e  solution given in book
Please have a look on that part
 [/INST] This is the same as the number of solutions in positive integers of:
$$5a+4w+3x+2y+z=20 \tag{1}, \\ \text{ where }w=b-a, x=c-b, y=d-c, z=e-d.$$
Equation $(1)$ tells us immediately that $a \in \{1, 2 \}$ and if $a=2$ there is only one possible solution.
Thus, we want to add $1$ to the number of possible solutions in positive integers of:
$$4w+3x+2y+z=15. \tag{2}$$
Again, $w \in \{1, 2 \}$.  If $w=2$, there is again only one possible solution.  Thus, we want to add $2$ to the number of possible solutions in positive integers of:
$$3x+2y+z=11. \tag{3}$$
Yet again, $x \in \{1, 2 \}$.  This time, if $x=2$ there are two possible solutions.  If $x=1$ there are three possible solutions.  Thus, Equation $(3)$ has $5$ possible solutions in positive integers.
There are, therefore, $7$ possible solutions to the Equation $(1)$.
I really have no idea how the book is doing the counting.
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