hey if anyone could help me out with this question that'd be great. and if anyone has perms and combs tips because i absolutely cannot do it. thanks
and i think this question is from SGS 2019 (probably prelims) but no solution lol.
View attachment 35581
ii)
The total number of ways to pick 3 points for vertices from both lines is (n+n-1)C3 = (2n-1)C3.
The invalid triangles are when 3 points are picked from l1 or 3 from l2.
Therefore number of triangles = (2n-1)C3 - nC3 - nC3 = (2n-1)C3 - 2 x nC3 (this method is similar to probability with complementary events where you do 1-probability of something not happening).
iii) (alpha)
X has to be a vertex, so we need 2 points, 1 from each line, to create the triangle. (n-1)C1 x (n-1)C1 = (n-1)^2.
iii) (beta)
I'm going to add two scenarios - when X is a vertex and when X is not a vertex. When X is a vertex, both lines always need 1 vertex so this is the answer from part (iii alpha), being (n-1)^2. Next scenario is when X is not a vertex. Picking two vertices on l1 gives (n-1)C2, then picking 1 on l2 is (n-1)C1 as we need to exclude point X. Therefore (n-1)^2 + (n-1)C2 x (n-1)C1 = (n-1)^2 + (n-1)^2(n-2)/2 = (n-1)^2 x (1 + n/2 - 1) = n/2 (n-1)^2.
iv)
The total number of triangles with two vertices on l2 are n/2 (n-1)^2 similar to part (iii beta). This is equal to the number of triangles (ii), minus the number of triangles with two vertices on l1 (iii beta), plus the number with X as a vertex (iii alpha) because minusing (iii beta) gets rid of the triangles with vertex X.
n/2 (n-1)^2 = (2n-1)C3 - 2 x nC3 - n/2 (n-1)^2 + (n-1)^2
n (n-1)^2 - (n-1)^2 + 2 x nC3 = (2n-1)C3
(2n-1)C3 = 2 x nC3 + (n-1)^2 (n-1)
(2n-1)C3 = 2 x nC3 + (n-1)^3
lemme know if some explanations didn't make sense my brain is lagging
EDIT: I agree with ur answer for (i) as nC3
