How many length-$k$ strictly decreasing sequences where sum is $N$? 
How many strictly decreasing sequences of length $k$ in positive integers can I find where the sum of elements is $N$?

The problem can be described this way too,
I have  a number $N$ . Now I want to divide $N $ into $k$ groups , so that group $i$ gets strictly more than group $(i+1)$ and so on .
For example, I have $N = 8 , k = 3$.
Then,
5 2 1 
and
4 3 1
So in this case, there are two answers.
 A: In order to transform the strictly decreasing partition into a normal partition, remove $k-1, k-2$ etc from each successive partition (note that nothing is removed from the last partition). This is then a normal partition, sorted into decreasing order (but not strict-decreasing). The total number removed is $k(k-1)/2$. We can then work with a partition of $n' = n-k(k-1)/2)$ into $k$ parts.
So, when partitioning into a fixed number of parts:

The number $p_k(n)$ of partitions of $n$ into exactly $k$ parts is equal to the number of partitions of $n$ in which the largest part has size $k$. The function $p_k(n)$ satisfies the recurrence
$$p_k(n) = p_k(n − k) + p_{k − 1}(n − 1)$$
  with initial values $p_0(0) = 1$ and otherwise $p_k(n) = 0$ if $n ≤ 0$ or $k ≤ 0$. 

Note also that $p_1(a) = p_a(a) = 1$ and $p_k(n) = 0$ when $k>n$

So, for your example, we have $n=8, k=3$, so $n'=n-k(k-1)/2 = 8-3 =5$. Then partitioning $n'=5$ into $k=3$ parts:
$$\begin{align}
p_3(5) &= p_3(2)+p_2(4)\\
&=0 + p_2(2)+p_1(3)\\
&=1+1 = \fbox{2}\\
\end{align}$$
A: An alternative way to consider is the following. Put
$N(k,n) =$ number of strictly decreasing (= increasing) sequences of length $k$ and sum $n$, with
$N(0,0)=1$.
and consider that
$M(q,n) =$ number of strictly decreasing (= increasing) sequences of length up to $q$ and sum $n$ =
= number of partitions of $n$ into distinct parts,  i.e.
$$ M(q,n) = \sum\limits_{0\, \leqslant \,j\, \leqslant \,q} {N(k,n)} $$
then $M(q,n)$ is given by being:
$$
\prod\limits_{1\, \leqslant \,j\, \leqslant \,q} {\left( {1 + x^{\,j} } \right)}  = \sum\limits_{0\, \leqslant \,n} {M(q,n)x^{\,n} } 
$$
(Refer for instance to
Lectures on Integer Partitions - Herbert S. Wilf )
from which the equivalent expression for $N(k,n)$ follows easily. 
Also, by considering the possible values of the last term in the sequence, it is easy to derive that $N(k,n)$ satisfies the recurrence
$$
\left\{ \begin{gathered}
  N(k,0) = \left\{ \begin{gathered}
  1\quad k = 0 \hfill \\
  0\quad 0 < k \hfill \\ 
\end{gathered}  \right. \hfill \\
  N(k,n) = \sum\limits_{0\, \leqslant \,j\, \leqslant \,n} {N(k - 1,n - j)} \quad \left| {\;1 \leqslant k} \right. \hfill \\ 
\end{gathered}  \right.
$$
