In how many ways can a positive integer $n$ be expressed as a summation of positive integers less than $n$ For example if I take $n=5$, then I can express it in the following ways:


*

*$1+1+1+1+1$

*$2+3$

*$3+2$

*$1+4$

*$4+1$

*$1+1+3$

*$1+3+1$

*$3+1+1$

*$2+2+1$

*$2+1+2$

*$1+2+2$


Please note that the order of terms in the expression of summation also counts if the terms are distinct.
 A: Take the string $ \large 1{\boxed.} 1\boxed. 1\boxed. 1\boxed. 1$
In the $\;(n-1)\;$ boxes, either put a $+\;$ or a comma.
$1,\;\;1,\;\;1+1+1$ e.g. would represent $1+1+3$
Since you have $2$ choices for each box, # of compositions = $2^{n-1}$
but as you have specified positive integers less than $n$, ans = $2^{n-1} - 1 = 2^4 -1 = 15$
A: If we count plain $n$ as a legitimate expression, there are $2^{n-1}$.
In general, let $a_n$ be the number of ways to decompose $n$. We show that $a_{n}=2^{n-1}$. The proof is by induction on $n$. Clearly $a_1=1$.
There are two types of compositions of $n+1$: (i) the ones that end with $1$ and (ii) the ones that end with a number $\ge 2$.
The Type (i) compositions of $n+1$ are obtained by appending a $1$ to a composition of $n$. By the induction hypothesis, there are $2^{n-1}$ Type (i) compositions of $n+1$.
The Type (ii) compositions of $n+1$ are obtained by adding $1$ to the last entry of a composition of $n$. There are $2^{n-1}$ of these. 
This gives a total of $2^n$, and completes the induction step.
Remark: If we break with tradition and do not allow plain $n$, the number of ways to decompose $n$ is $2^{n-1}-1$. 
A: We can also do this by finding the number of ordered partitions of $n$ into exactly $r$ parts and then summing them over from $1$ to $n$.
When we have exactly $r$ partitions of $n$, we have the generating function $(x+x^2+x^3+...)^r=x^r/(1-x)^r$ 
Now, we must pick the coefficient of $x^n$ from
$x^r[1+\binom{r}{1}x + \binom{r+1}{2} x^2 +.....+ \binom{r+(n-r-1)}{n-r}x^{n-r}+....]$
which is  $\binom{n-1}{r-1}$
This is the general expression for orders partition of $n$ into $r$ parts.
To find all partitions we must evaluate 
$\sum_{i=1}^n \binom{n-1}{r-1} = \binom{n-1}{0} + \binom{n-1}{1} + \binom{n-1}{2} +....+ \binom{n-1}{n-1}$
Considering the binomial expansion of $(1+x)^{n-1}$, we have 
$\binom{n-1}{0} + \binom{n-1}{1}x+ \binom{n-1}{2}x^2 +....+ \binom{n-1}{n-1}x^{n-1}$
Putting $x=1$ in the above expression,we get
$\sum_{i=1}^n \binom{n-1}{r-1} =(1+1)^{n-1}=2^{n-1}$
But this also contains the partition of the form $n=n$, which is not allowed.So required number of partitions is $2^{n-1} -1$.

For this question with $n=5$, total number of required partitions is 

 $2^{5-1}-1=2^4-1=15$

