How many ways are there of sum $\frac{n(n+1)}{2}$ with $n$ addends? Knowing than $i$ appears at most $n+1-i$ times.
I really haven't had any important ideas. When speaking of a fixed number of addends, we leave out the partitions of a number. How can I proceed; any suggestions?
On
HINT
It is not clear whether you are considering the Partitions
or Compositions of $s=\binom{n+1}{2}$.
In the first case we have the $n$ addenda ordered non-decreasingly (or non-increasingly).
The corresponding generating function is clearly
$$
\eqalign{
& P(x,n) = \cr
& = \left( {1 + x} \right)^n \left( {1 + x^{\,2} } \right)^{n - 1} \cdots \left( {1 + x^{\,n - 1} } \right)^2 \left( {1 + x^{\,n} } \right)^1 = \cr
& \prod\limits_{k = 1}^n {\left( {1 + x^{\,k} } \right)^{n + 1 - k} } \cr}
$$
In the second case the addenda are $n$-tuples (vectors), where the order matters. The generating function is then $$ \eqalign{ & Q(x,n) = \cr & = \left( {x + x^{\,2} + \cdots + x^{\,n} } \right)\left( {x + x^{\,2} + \cdots + x^{\,n - 1} } \right)\; \cdots \;x \cr & = x^{\,n} \prod\limits_{k = 1}^n {\left( {{{1 - x^{\,k} } \over {1 - x}}} \right)} = \left( {{x \over {1 - x}}} \right)^{\,n} \prod\limits_{k = 1}^n {\left( {1 - x^{\,k} } \right)} \cr} $$
Let $N=\frac{n(n+1)}{2}$. We know that $$N=\sum_{i=1}^{N}1=1+1+...+1$$
Notice that in this expression there are $N$ "ones" and $(N-1)$ "plus signs". Then we can choose $n-1$ "plus signs" to keep fixed, and collapse the rest by computing the sums (see at bottom for an example). The number of ways to do this is $${ N-1\choose n-1}={ \frac{n(n+1)}{2} - 1\choose n-1}$$ Note this does not adjust for repetition.. so for example $1+2$ and $2+1$ would be distinct
To adjust for repetition you can build off this answer.
Example: Say we want to write 4 as a sum of 2 numbers. Then $$4=1+1+1+1$$ which has 3 "plus signs". we can fix the first, second or third plus sign, and collapse the non-fixed ones like so: $$(1)\underline{+}(1+1+1) =1+3$$ $$(1+1)\underline{+}(1+1) =2+2$$ $$(1+1+1)\underline{+}(1) = 3+1$$