Question

Points A, B, C, D, E, F, G, H and I are arranged in two rows as shown on the right.
How many triangles can be formed having vertices chosen from these points?
How many of these triangles have point A as one of their vertices?


I thought the answers would be 7 triangles, 1 triangle but the solutions in the book say 70 triangles, 22 triangles. Why is this? Thank you :)

Answer 1

Well, if \(A\) is a vertex, then we can focus on the other two vertices.

Answer 2

The second vertex can be B,C,D or E.
The third vertex can be F,G,H or I.

Answer 3

Hmmm I've tried several things but still didn't reach 70 triangles or 22. This question comes under the topic of combinations - would you please be able to start the question off?

Answer 4

if i do by manual, i got 22 ways
but i cant get the formulation of combinatoric

Answer 5

btw, it just for number b have a start for point A

Answer 6

I have worked on this problem for a long time. I am now at 63 for part A although I was at 74. Mindboggling that such a counting problem could be so vexing.

Look at the info at the link below for practically the same problem with 5 of the points in a row (line segment) above the other 4 points. The answer there is also 70. (I think the answer might be the same if one point were on the top row with eight on the bottom row.) While I can follow the arguments made for 5 on top and 4 on bottom, I cannot get 70 for the problem posted here.

What is cool about this link is that several strategies are given for the solution. Do take a look: http://tinyurl.com/cpdbh49

Answer 7

@sedighn

Answer 8

I think it should be 70 for the first part because....
For forming Triangles ,
You can choose from
(1) row 1st two points, that is , in 5C2 ways and choose 1 point out of 4 points of 2nd row in 4C1 ways... So first case is. 5C2* 4C1 = 40 triangles
(2) row 2nd two points, that is in 4C2 ways and choose the 1 point out of 5 points of 1st row in 5C1 ways... So 2nd case is 4C2*5C1 = 30 triangles

In total 40+30 = 70 triangles...

Answer 9

For the second part....
again two cases... as vertex A is fixed,
(1) you can choose 1 point from 4 points left in 1st row and 1 point from 4 points of 2nd row ==> 4C1 * 4C1 = 16 triangles
or
(2) you can choose both the points from 2nd row ==> 4C2 = 6 triangles

In total 16+6 = 22 triangles :-)

Answer 10

Thank you so much guys for taking the time :)