number of ordered partitions of integer How to evaluate the number of ordered partitions of the positive integer $ 5 $?
Thanks!

Since $5$ is a smallish number, it is reasonable to try to list all of the ordered partitions, and then count.  First maybe, lest we forget, write down the trivial partition $5$.  Then write down $4+1$, $1+4$. Now list all the ordered partitions with $3$ as the biggest number. This is easy, $3+2$, $2+3$, $3+1+1$, $1+3+1$, $1+1+3$.  Continue.  After not too long, you will have a complete list. 
It so happens that for this type of problem, there is a simple general formula, which one might guess by carefully finding the number of ordered partitions of $1$, of $2$, of $3$, of $4$.  And there are good ways of proving that the general formula holds.  Let us deal with the case $n=5$.
Put $5$ pennies in a row, leaving a little gap between consecutive pennies.  There are $4$ interpenny gaps.  CHOOSE any number of these gaps ($0$, $1$, $2$, $3$, or $4$) to put a grain of rice into.  Any such choice gives rise to a unique ordered partition of $5$, and all of them arise in this way. For example, the trivial partition $5$ comes from using no grain.  The partition $4+1$ comes from putting a grain of rice after the $4$th penny. And so on.  So there are exactly as many ordered partitions of $5$ as there are ways of choosing a SUBSET of the set of gaps.  But a set of $4$ elements has $2^4$ subsets.
Or else one could attack the problem by induction.  For example, let $P(n)$ be the number of ordered partitions of $n$.  Now look at $P(n+1)$.  Ordered partitions of $n+1$ are of two types: (i) last element $1$ and (ii) last element bigger than $1$.  You should be able to see that there are $P(n)$ ordered partitions of $n+1$ of each type, meaning that $P(n+1)=2P(n)$.
But after all this fancy stuff, I would like to urge that you get your hands dirty, that you list and count the ordered partitions of $n$ for $n=1$, $2$, $3$, $4$, $5$, maybe even $6$.  
