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?

  • $\begingroup$ I think you are entirely correct... but I am not an expert. :) So if you don't need the answer urgently then I suggest you wait for a more expert voice as well. $\endgroup$ – antkam Apr 8 at 22:37

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