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Also asked on overflow: https://mathoverflow.net/questions/296567/some-elementary-schubert-calculus-calculations/296583#296583

Consider $3$ dimensional projective space (although you don't have to know anything specific, just that there is no parallelism.) Here is the question:

How many pairs of intersecting lines in space are there such that:

1) Their plane passes through a given fixed line $l$.

2) Their point of intersection lies on a given plane $P$.

3) Together they intersect $4$ given lines $l_1,l_2,l_3,l_4$ ?

The answer in the book listed is $17$, assuming there is no typo. The following is useful.

$\bf{Theorem}$ Given four lines in space, there are $2$ lines intersecting all four.

${\bf Thoughts :}$ Well, you don't have to read this if you want to approach the question fresh, but I can tell you where my problem lies, so you get the core quicker.

First, either one of the lines intersects $3$ of $l_1,l_2,l_3,l_4$, or each line intersects $2$ of them.

In the first case there are $\binom{4}{3}=4$ many triples of lines from $l_1,l_2,l_3,l_4$, say $l_1,l_2,l_3$, and together with $l$ there are two lines intersecting all four. Each of these intersects $P$ is a point. Now all that is needed is a line through this point and the lines $l$ and $l_4$, this is unique as is easy to see. This gives $2\times 4=8$ line pairs.

In the second case each line must intersect $2$ of the lines $l_1,l_2,l_3,l_4$. There are $\frac{1}{2}\binom{4}{2}=3$ many ways to divide this. Now if the book is right for each such division there must be $3$ solutions. This is were I cannot proceed further.

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