Let $n_1<n_2<n_3<n_4<n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5=20$.  Then the number of such distinct arrangements $\left(n_1,n_2,n_3,n_4,n_5\right)$is 

$1^{\text{st }}$solution - As $n_1\ge 1,n_2\ge 2$ etc.|
We have $a+b+c+d+e=20-(1+2+3+4+5)=5$
i.e. $a+b+c+d+e=5$
Now $a\le b\le c\le d\le e.$ So the cases can be listed giving 
$\begin{array}{l}(0,0,0,0,5);(0,0,0,1,4);(0,0,0,2,3);(0,0,1,1,3);\\ (0,0,1,2,2);(0,1,1,1,2);(1,1,1,1,1)\end{array}$
So there are $7$distinct arrangements.

$2^{\text{nd }}$Solution - The largest e is $10$. (given)
$$\begin{gathered} a b c d e \\ 1 2 3 4 10 \end{gathered}$$
 and least e is $6$.
$2,3,4,5,6$ 
So the number of ways ${=}^4{C}_0+\frac{{ }^4{C}_1}{4}+\frac{{ }^4{C}_2}{3}+\frac{{ }^4{C}_3}{2}+\frac{{ }^4{C}_4}{1}$
                                        $=1+1+2+2+1=7$