can't do Perms and Combs (1 Viewer)

english? ensquish

New Member
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.

english? ensquish

New Member
i thought i) might be like nC3 but i'm never right and i don't know about the rest :'p

pikachu975

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

english? ensquish

New Member
(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.
Thank you so much for your explanations.

I didn't get this section of the question where you changed (n-1)C2 x (n-1)C1 to (n-1)^2(n-2)/2