How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? 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 $(n_1, n_2, n_3, n_4, n_5)$ is......
I have no idea how to proceed. Manually, I have done it
$$1+2+3+4+10$$
$$1+2+3+5+9$$
$$1+2+3+6+8$$
$$1+2+4+5+8$$
$$1+2+4+6+7$$
$$1+3+4+5+7$$
$$2+3+4+5+6$$
But is there any way I can do it by Permutation and Combination method?

A variation based upon generating functions. We introduce positive integers $a,b,c,d$ and put
\begin{align*}
n_2&=n_1+a\\
n_3&=n_2+b=n_1+a+b\\
n_4&=n_3+c=n_1+a+b+c\\
n_5&=n_4+d=n_1+a+b+c+d
\end{align*}
The equation $n_1+n_2+n_3+n_4+n_5=20$ transforms to
\begin{align*}
5n_1+4a+3b+2c+d=20\tag{1}
\end{align*}
with $n_1,a,b,c,d>0$.

In order to find the number of solutions of (1) we consider the generating function $A(x)$
  \begin{align*}
A(x)&=\frac{x^5}{1-x^5}\cdot\frac{x^4}{1-x^4}\cdot\frac{x^3}{1-x^3}\cdot\frac{x^2}{1-x^2}\cdot\frac{x}{1-x}\\
&=x^{15}+x^{16}+2x^{17}+3x^{18}+5x^{19}+\color{blue}{7}x^{20}+10x^{21}+\cdots
\end{align*}
  and obtain with some help of Wolfram Alpha the solution
  \begin{align*}
[x^{20}]A(x)\color{blue}{=7}
\end{align*}

Add-on: Some  details
We first  transform the  equation with restrictions by introducing positive integers $a,b,c,d$ in an equivalent equation with more convenient restrictions
\begin{align*}
&n_1 + n_2 + n_3 + n_4 + n_5 = 20\qquad&\qquad&5n_1+4a+3b+2c+d=20\\
&0<n_1<n_2<n_3<n_4<n_5\qquad&\qquad&0<n_1,0<a,0<b,0<c,0<d
\end{align*}
We now consider admissible $5$-tuples $(n_1,a,b,c,d)$. Increasing $n_1$ by $1$ adds $5$ to the equation. Similarly, increasing $a$ by $1$ adds $4$ to the equation. We encode these increments via exponents of generating functions:


*

*$n_1$: Increment by $5$ gives
\begin{align*}
x^5+x^{10}+x^{15}+\cdots=x^5(1+x^5+x^{10}+\cdots)=\frac{x^5}{1-x^5}
\end{align*}

*$a$: Increment by $4$ gives
\begin{align*}
x^4+x^8+x^3+\cdots=x^4(1+x^4+x^8+\cdots)=\frac{x^4}{1-x^4}
\end{align*}


and similarly for $b,c$ and $d$. Observe that each of $n_1,a,b,c,d$ is positive, i.e. has at least value $1$. This is respected by smallest values $x^5,x^4,x^3,x^2$ and $x^1$.

The number of admissible solutions is therefore
  \begin{align*}
[x^{20}]&\frac{x^5}{1-x^5}\cdot\frac{x^4}{1-x^4}\cdot\frac{x^3}{1-x^3}\cdot\frac{x^2}{1-x^2}\cdot\frac{x}{1-x}\\
&=[x^{20}]\frac{x^{15}}{(1-x^5)(1-x^4)(1-x^3)(1-x^2)(1-x)}\\
&=[x^{5}]\frac{1}{(1-x^5)(1-x^4)(1-x^3)(1-x^2)(1-x)}\tag{2}\\
&=[x^{5}](1+x^5)(1+x^4)(1+x^3)(1+x^2+x^4)(1+x+x^2+x^3+x^4+x^5)\tag{3}\\
&=\cdots\tag{4}\\
&\color{blue}{=7}
\end{align*}

Comment:


*

*In (2) we use the coefficient of operator rule: $[x^{p}]x^qA(x)=[x^{p-q}]A(x)$.

*In (3) we expand the geometric series restricted to powers less or equal to $x^5$ since other terms do not contribute to $[x^5]$.

*In (4) we expand further and can omit terms with powers greater than $5$.
Hint: Instructive examples can be found in H.S. Wilf's book generatingfunctionology.
