If $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20.$ Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$ 
Let $n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$ be the positive integers such that $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20$
Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$ is

$\bf{My\; Try::}$ Let $n_{1} = x+1\;,n_{2} = y+2,n_{3}=z+3,n_{4}=t+4\;,n_{5}=u+5$
where $x,y,z,t,u\geq 0$ and $x\leq y \leq z\leq t \leq u$
So our equation convert into $x+y+z+t+u=5$
Now Put $x=0\;,$ We get $y+z+t+u=5\;,$ Then put $y=0\;,$ We get $z+t+u=5$
But using that way our task is very Complicated.
So plz explain me a better way to solve that equation.
Thanks
 A: Consider it as a problem where you have five boxes - $n_1,n_2,n_3,n_4,n_5$, and $5$ balls to be put into them. There are five cases -  (i) Only one box is used $([5])$ (ii) Two boxes are used $([4,1],[3,2])$ (iii) Three boxes used $([3,1,1],[2,2,1])$(iv)Four boxes used $([2,1,1,1])$ (v) All boxes used $([1,1,1,1,1])$  Adding them up, we get $7$.
A: Note that as 
$$n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$$
$$\implies n_1\geq 1,n_2\geq2 ...  n_5\geq 5$$
let 
$$n_1-1=x_1,n_2-2=x_2....,n_5-5=x_5$$
Hence, $$x_1+x_2+x_3+x_4+x_5=5$$ and 
$$x_{1}\leq x_{2}\leq x_{3} \leq x_{4} \leq x_{5}$$
These are pretty tame inequalities under these restrictions, now I don't know how to make a table here, but I get $7$ possible cases.
A: Here is an elementary argument.
The minimum possible $n_1$ is $1$. 
Since the sum is $20$, dividing $20$ by $5$ we get $4$, which means $n_1$ cannot be $4$, since the numbers are strictly increasing. 
If $n_1=3$, increasing each following number by $1$, we get $3+4+5+6+7=25$. 
So the maximum possible $n_1$ is $2$. With $2+3+4+5+6=20$ we get the only possible arrangement when $n_1=2$.
Now come back to $n_1=1$. We consider what is the maximum possible $n_2$. Using the same argument as above, we see that $4$ does not work. $n_2=3$ only gives one arrangement $1+3+4+5+7=20$. 
So now $n_1=1, n_2=2$. We consider the maximum possible $n_3$. We find that it is $4$.
Now two cases: $n_1=1, n_2=2, n_3=4$ and $n_1=1, n_2=2, n_3=3$. In each case you can discuss as above what is the maximum possible $n_4$. With the first case, we get $1+2+4+6+7$ and $1+2+4+5+8$. With the second case we get other $3$ arrangements.
