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: 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.
A: Write
$$n_1=1+y_1,\qquad n_k=n_{k-1}+1+y_k \quad(2\leq k\leq5)$$
with $y_k\geq0$ $(1\leq k\leq 5)$. Collecting terms we then obtain
$$20=\sum_{k=1}^5 n_k=15 + 5y_1+4y_2+3y_3+2y_4+y_5\ .$$
We therefore have to count the solutions of
$$\sum_{k=1}^5 z_k\,k=5$$
in integers $z_k=y_{6-k}\geq0$. Each such solution encodes a partition of $5$ into $z_k$ parts of size $k$. Since there are $7$  partitions of $5$, the answer to the original question is $7$.
A: Numerical algorithm: 
Let $S_{m,k}$ count the solutions of $n_1 + n_2 +\cdots + n_k=m$ with $n_1 < n_2 \cdots < n_k$
Let $T_{m,k,t}$ be the same, subject to $n_k=t$. 
Then  $$T_{m,k,t}=\sum_{s=1}^{t-1} T_{m-t,k-1,s}$$
And $S_{m,k}=\sum T_{m,k,t}$. Together with the bondary conditions, this allows to compute $S_{m,k}$
For example (Java, non optimized) https://ideone.com/BZjsmQ
Gives $S(20,5)=7$
A: Let $m_1 = n_1, m_2 = n_2 -1, m_3 = n_3 -2, m_4 = n_4 -3, m_5 = n_5-4$; then $m_1 \leq m_2 \leq \cdots \leq m_5$ and $m_1+m_2+m_3+m_4+m_5 = 10$. Thus we need the number of 5 partitions of 10, $P(10,5)$. Clearly, $P(10, 5) = 7$, using the recurrence $P(n,p) = P(n-1, p-1) + P(n-p,p)$.
A: You could start with $1+2+3+4+5 = 15$ and see that you still have to add $5$.
Adding to e.g. $n_3$ implies that you also have to add to $n_4$ and $n_5$, so you need $3$ to do that addition.
So finally you'll end up with $5n_1 + 4n_2 + 3n_3 + 2n_4 + n_5 = 5$.
Which you could try to solve with recursion, either with a program or with a formula. I'm not sure if the formula will be easy and closed form.
A: Using combinatorics  we find that the answer is coefficient of $x^{20} $ in $(x+x^2+x^3+x^4+x^5)(x^2+x^3+..x^6)(x^3..+x^7)(x^4+..+x^8)(x^5+..+x^9)=x^{15}(1+x+x^2+x^3+x^4)^5$  which is $7$ ie ways are $(1,1,1,x,x^4), (1,1,x,x,x^3), (1,1,1,x^2,x^3), (x,x,x,x,x), (1,1,x,x^2,x^2),(1,x,x,x,x^2), (1,1,x,x,x^3) $
