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

For when $e=10$, there is only one arrangement of numbers that satisfies $a+b+c+d+e=20$, which is $(1,2,3,4,10)$. I am going to call this solution our default solution.
Consider when $e=9$, there are four different ways to make sure that the numbers sum to 20. You can choose one number out of $a,b,c,d$ from the default solution and add one to it. The solutions for $e=9$ are the following: $(2,2,3,4,9)$, $(1,3,3,4,9)$, $(1,2,4,4,9)$, $(1,2,3,5,9)$. There is only one solution here which fulfills $a<b<c<d<e$ and that solution is $(1,2,3,5,9)$.
Note that for $e=9$, there are $4\choose1$ possibilities since we pick any one out of $a,b,c,d$ to increase, but there is only one correct answer.
For $e=8$, there are two different ways to increase numbers in the default solution. You could either (a) pick two numbers out of $a,b,c,d$ and increase them both by 1, or you could (b) pick one number out of $a,b,c,d$ and increase that number by 2.
For (a), the possibilities are:
$(2,3,3,4,8)$
$(2,2,4,4,8)$
$(2,2,3,5,8)$
$(1,3,4,4,8)$
$(1,3,3,5,8)$
$(1,2,4,5,8)$.
For (b), the possibilities are:
$(3,2,3,4,8)$
$(1,4,3,4,8)$
$(1,2,5,4,8)$
$(1,2,3,6,8)$.
From (a), the only solution is $(1,2,4,5,8)$. From (b), the solution is $(1,2,3,6,8)$. Notice that for $e=8$, there are $4\choose2$ and $4\choose1$ possibilities for (a) and (b) respectively.
Following this pattern, for $e=7$ there will be ${4\choose3}+{4\choose2}+{4\choose1}$ different possibilities corresponding to how many numbers you are changing. For $e=6$ there are ${4\choose4}+{4\choose3}+{4\choose2}+{4\choose1}$ possibilities.
To reconstruct the equation you gave, I believe you have to count the instances $n$ which each $4\choose k$ occurs and divide by $n$. For example, $4\choose{k=1}$ occurs $n=4$ different times so that term in the equation is $\frac{{4\choose1}}{4}$. In total, there are 4 instances of $4\choose1$, 3 instances of $4\choose2$, 2 instances of $4\choose3$, and 1 instance of $4\choose4$. This results in the equation $\frac{{4\choose1}}{4}+\frac{{4\choose2}}{3}+\frac{{4\choose3}}{2}+\frac{{4\choose4}}{1}$. For the final $4\choose0$ term, I think that would be the scenario where $e=10$ since that is the default solution and we pick 0 numbers to change.
