Consider a finite projective plane of order 11, which is there more of:
a) Unordered 4-tuples of lines with a non-empty intersection?
b) Unordered 7-tuples of points which belong to the same line?
Now what I got is:
$|X|= |L| = n^2 + n + 1$ So there are exactly 133 points and 133 lines.
Because the plane is of order 11 i know that there are 12 points on a single line.
a) Each two lines intersect at only one point, so for 4 lines to have a non-empty intersection they'd have to intersect in the same point.
For each point I know that there are 12 lines which interesect it so I get: $$133 \cdot \binom{12}{4} $$
b) Each line consists of 12 points so to get unordered 7-tuple of points $$133\cdot\binom{12}7{}$$
Therefore there are more unordered 7-tuples of points belonging to the same line than there are 4-tuples of lines inciding with the same point.
Is this correct or did I do something wrong?