Number of distinct arrangements {$n_i$} $n_1Let  $n_1<n_2<n_3<n_4<n_5$  be positive integers such that $n_1+n_2+n_3+n_4+n_5=20$.Then what is the number of such distinct arrangements $(n_1,n_2,n_3,n_4,n_5)$?
My Approach : 
I assumed 
$n_1=t_0+1$
$n_2=n_1+t_1+1$
$n_3=n_2+t_2+1$
$n_4=n_3+t_3+1$
$n_5=n_4+t_4+1$
Where $t_0,t_1,t_2,t_3,t_4 \ge 0$
Now the sum becomes :
$5t_0+4t_1+3t_2+2t_3+t_4=5$
After this, I put the values of $t_i$s  $0,1,..$ and so on, and therefore found $7$ Solutions.
My question : Is there another way to solve this question, because as this question was asked in a competitive exam (JEE Advanced), This a very long solution. 
 A: I would say that you start with $1+2+3+4+5=15$ and have to distribute 5 more. And the 5 more have to satisfy the rule that you can't distribute more to the $i$th position than the $i+1$th position.
From that I can come up with the solutions for distributing those 5 faster than I can write them down.  They are obviously $(1,1,1,1,1), (0,1,1,1,2), (0,0,1,1,3), (0,0,1,2,2), (0,0,0,1,4), (0,0,0,2,3), (0,0,0,0,5)$ and I count them.
This is obviously a one-off trick and that is appropriate to the actual values chosen.  However the described rule comes down to counting the number of partitions of 5.  See https://en.wikipedia.org/wiki/Partition_(number_theory) for more on partitions.
As for your harder question with $n_1+n_2+n_3+n_4+n_5=50$, that can be calculated recursively.  You want the number of partitions of $35$ into no more than $5$ groups.  Well, define $p_{i,k}$ to be the number of partitions of $i$ into no more than $k$ groups.  We have the following rules:


*

*$p_{0,k} = 1$ (just a string of 0s)

*$p_{i,1} = 1$ (all go into one)

*$p_{i,k+1} = p_{i,k} + p_{i-k,k+1}$ (add one to everything)


From here we can work out rules like:
$p_{i,1} = 1$
$p_{i,2} = i+1$
$p_{i,3} = 1 + 2 + ... + (i+1) = (i+1)(i+2)/2$
And so on until we have a formula to use.  But finishing that would be harder and not appropriate for the place it appeared. :-)
A: We can use casework. It is important to be organized to make sure you don't miss any cases.


*

*Smallest is 1 and then 2: 


*

*$\{1, 2,  3, 4, 10\}$

*$\{1, 2,  3, 5, 9\}$

*$\{1, 2,  3, 6, 8\}$

*$\{1, 2,  4, 5, 8\}$

*$\{1, 2,  4, 6, 7\}$

*We cannot have repeats so that is all


*Smallest is 1 and then 3: 


*

*$\{1, 3,  4, 5, 7\}$


*Smallest is 2


*

*$\{2, 3, 4, 5, 6\}$



Total, there are $\boxed{7}$ solutions. Surprisingly easy for an JEE problem. This can be solved much faster if you just start listing all numbers.
