Find the number of distinct arrangements of 5 unequal positive integers such that their sum is 20 Let the integers be $n_1, n_2, n_3, n_4$ and $n_5$.
It's given in the question that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. 
I thought of taking $n_2$ as $n_1 + a$, $n_3$ as $n_1 + a + b$ and so on... where a, b,... are not equal to 0.
So I got this expression:
$5n_1 + 4a + 3b + 2c + d = 20$
After this, I'm not able to continue. How do I proceed?
Thanks in advance.
 A: Integer partitions of $n$ into $k$ parts is given by the recurrence
$$p(n,k)=p(n-1,k-1)+p(n-k,k)\tag{1}\label{1}$$
Since a partition of $n$ into $k$ parts either has $1$ as it's smallest part in $p(n-1,k-1)$ ways or it has it's smallest part greater than $1$ in $p(n-k,k)$ ways. This recurrence has $p(n,1)=1$ and $p(1,1)=1$ and $p(1,k)=0$ for $k\gt 1$.
If, say, the integer partition is
$$\sum_{r=1}^{k}n_r=n$$
such that $n_1\le n_2\le \cdots\le n_k$ we make the substitution $n_r'=n_r+(r-1)$ then we have
$$\sum_{r=1}^{k}n_r'-\sum_{r=1}^{k}(r-1)=n$$
$$\implies\sum_{r=1}^{k}n_r'=n+\binom{k}{2}=n'$$
for $n_1'\lt n_2'\lt\cdots\lt n_k'$ this is a bijection between integer partitions of $n$ into $k$ parts and integer partitons of $n'$ into $k$ distinct parts.
In other words the number of partitions of $n'$ into $k$ distinct parts is equal to the number of partitions of $n=n'-\binom{k}{2}$ into $k$ parts.
$$p_d(n',k)=p(n'-\text{C}(k,2),k)$$
In your case $n'=20$ and $k=5$ and $\binom{5}{2}=10$ so $n=20-10=10$.
Reading off $p(10,5)$ in the table formed by the recurrence $\eqref{1}$
$$\begin{array}{cc}
&k\\
n&\begin{array}{c|cccccccccccc}
 p(n,k)&1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 &12\\\hline
1 &1&&&&&&&&&&&\\           
2 &1 &1&&&&&&&&&&\\          
3 &1 &1 &1&&&&&&&&&\\         
4 &1 &2 &1 &1&&&&&&&&\\        
5 &1 &2 &2 &1 &1&&&&&&&\\       
6 &1 &3 &3 &2 &1 &1&&&&&&\\      
7 &1 &3 &4 &3 &2 &1 &1&&&&&\\     
8 &1 &4 &5 &5 &3 &2 &1 &1&&&&\\    
9 &1 &4 &7 &6 &5 &3 &2 &1 &1&&&\\   
10 &1 &5 &8 &9 &\bbox[#FFA,10px]{7} &5 &3 &2 &1 &1&&\\  
11 &1 &5 &10 &11 &10 &7 &5 &3 &2 &1 &1&\\ 
12 &1 &6 &12 &15 &13 &11 &7 &5 &3 &2 &1 &1\\
\end{array}\end{array}$$
Thus, we have
$$p_d(20,5)=p(10,5)=7\tag{Answer 1}$$
If you want the order of the parts to count then multiply that by $5!$ to give
$$7\cdot 5!=840\tag{Answer 2}$$
