Let $2\leq k\leq r\leq n$ are positive integers and $r=kt$.
I construct sets such that $\cup\_{i=1}^n A\_i=\{1,2,3,\dots,n\}=X$, this union is disjoint and if $x\in A\_i$ and $y\in A\_j$ for all $i\leq j$, then $x<y$. I put one condition such that any $t$ sets will contain $k$ elements and the others will be single elements. How can I count the number of such sets?


Example: Let $n=8$, $r=6$, $k=2$ and since $r=kt$, $6=23$, $t=3$.
then
We can spilts $9$ different ways with above conditions as in the following:
$$1=\{A\_1=\{1,2\},A\_2=\{3,4\},A\_3=\{5,6\},A\_4=\{7\},A\_5=\{8\}\}$$
$$2=\{A\_1=\{1,2\},A\_2=\{3,4\},A\_3=\{5\},A\_4=\{6,7\},A\_5=\{8\}\}$$
$$3=\{A\_1=\{1,2\},A\_2=\{3,4\},A\_3=\{5\},A\_4=\{6\},A\_5=\{7,8\}\}$$
$$4=\{A\_1=\{1,2\},A\_2=\{3\},A\_3=\{4,5\},A\_4=\{6,7\},A\_5=\{8\}\}$$
$$5=\{A\_1=\{1,2\},A\_2=\{3\},A\_3=\{4,5\},A\_4=\{6\},A\_5=\{7,8\}\}$$
$$6=\{A\_1=\{1,2\},A\_2=\{3\},A\_3=\{4\},A\_4=\{5,6\},A\_5=\{7,8\}\}$$
$$7=\{A\_1=\{1\},A\_2=\{2,3\},A\_3=\{4,5\},A\_4=\{6,7\},A\_5=\{8\}\}$$
$$8=\{A\_1=\{1\},A\_2=\{2,3\},A\_3=\{4,5\},A\_4=\{6\},A\_5=\{7,8\}\}$$
$$9=\{A\_1=\{1\},A\_2=\{2,3\},A\_3=\{4\},A\_4=\{5,6\},A\_5=\{7,8\}\}.$$



Combinatoric Problem
Because $t$ sets contain $k$ elements each, there are $n-kt$ singletons and $t+n-kt$ sets overall. Each partition is completely determined by choosing the $t$ sets, and there are $\binom{t+n-kt}{t}$ ways to do that.
For your example, this is
$$\binom{3+8-6}{2}=\binom{5}{2}=10.$$
You are missing the one with $A\_1=\{1\}$ and $A\_2=\{2\}$.


