How many combinations of intersections of n hyperplanes there are in which a hyperplane is not crossed more than k times? Consider you have $n$ hyperplanes with dimension $k$ that are not coplanar. Each hyperplane intersects the others $n-1$ times. Any intersection of $k$ such hyperplanes produces a vector. The number of such intersections is therefore
$$d=\left(\begin{array}{c}
n\\
k
\end{array}\right)$$
From these $d$ intersections, I am interested in choosing $n$ intersections, but only those combinations that will have any hyperplane intersected in exactly $k$ positions.
Here is an example with $n=4$ and $k=2$:

The lines represent the planes and the dots represent the intersections of the planes. The top row shows valid combinations, while the bottom row shows invalid ones.
I would like to calculate how many valid combinations exist, for any $k$ and $n>k$.
 A: I added the tag combinatorics and am doubtful about the tag arithmetic-geometry because the particular hyperplane arrangement is irrelevant. The question is simply this: given a set of $n$ symbols (which may as well be $\lbrace 1,2,\cdots,n\rbrace$) find $n$ (different) subsets of size $k$ so that every symbol is in exactly $k$ of the subsets. This could be called a symmetric incomplete block design $(v,b,r,k)=(n,n,k,k)$ but that term is usually reserved for symmetric balanced incomplete block design where we require each pair of symbols to be in $\frac{k(k-1)}{n-1}$ common blocks. 
later computing very small counts and using the OEIS finds information for the case $k=2$ and some for the case $k=3$.
In the special case that $k=2$ it is not hard to see that the number of ways for $n=3,4,5,6,7$ are $1,{\LARGE3},12,70,465.$ That is enough to locate  this sequence in the OEIS. There you will find references, a recurrence relation, asymptotics and the exponential generating function.  Also the concept of a frame:

[N]umber of 'frames' on n lines: given n lines in general position (none parallel and no three concurrent), a frame is a subset of n of the C(n,2) points of intersection such that no three points are on the same line.

The answer can be determined fairly quickly for any particular $n$  as a sum over various partitions (all parts at least $3$) of a product of multinomial coefficients (with some extra factorials in the quotient). evidently, all this can be encoded into an exponential generating function.   
details: When $k=2$ one could view the chosen blocks to be $n$ of the $\binom{n}{2}$ edges of the complete graph $K_n$ chosen so that each node is on two edges. This is called a $2$-factor and consists of one or more disjoint cycles with combined length $n$. With only one cycle there are $\frac{(n-1)!}{2}$ solutions. This is all for $n=3,4,5.$ When $n=6$ there are $\frac{5!}{2}=60$ one cycle solutions along with $\binom{6}{3}\frac{1}{2!}\cdot 1\cdot 1=10$ solutions with two $3$-cycles. For $n=7$ we have $360$ one cycle solutions along with $\binom{7}{4}  \frac{(3-1)!}{2} \frac{(4-1)!}{2}=105$ ways to separate the symbols into groups of size $3$ and $4$ and pick a cycle for each group.
For $n=11,k=2$ the "cycle-type" could be $11$ or $8,3$ or $7,4$ or $6,5$ along with $3,3,5$ and $3,4,4$. In general (for $k=2$) one can give an expression as a summation over the "cycle-types" of product/quotients of multinomial coefficients and factorials. 
LATER For the case $k=3$  the number of ways for $n=4,5,6$ are $1,12,330$ That is enough to locate the sequence  number of 3-regular 3-hypergraphs on n labeled vertices in the OEIS. A generating function (or at least a huge differential equation satisfied by one) and (complicated) recurrence are given. The usefulness seems limited as the numbers are found only out to $n=15$.  In retrospect that would have been a reasonable name to search for. In general your question could be phrased as:



How many $k$-uniform $k$-regular hypergraphs are there on $n$ labelled vertices?



