If 4 points are indicated on a line and

[ShyT]

If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?

a)20
b)30
c)40
d)70
e)90

answer is D.

I immediately recognized that it was a combinatronics problem but couldn't seem to rationalize a formula in my head. I initially tried drawing the 2 lines to see what kind of relationship I can get. I chose the top to be the 5 pointed line and the bottom to be 4 pointed line. I decided draw the triangle from top to botto m (1 point top 2 points bottom) I came to the formula that for every 1 point on top there will be 4!/2!2! on bottom so 6...and for total # triangles on top 6X5=30 (triangles possible w 1 point on top 2 points bottom) and then flipped it to do the same thing 5!/2!3!=10 x 4=40 30+40=70. Even though this makes sense to me now it took me forever to rationalize it. Is there something that I am not seeing that should trigger a quicker response? Thanks!

[RonPurewal]

hi,

this folder is only for problems from the OFFICIAL GMAT PREP SOFTWARE.
is this problem from that software?

* if so, please post a screenshot, either directly or by uploading to an image hosting site such as postimage.org.

* if not, then, unfortunately, we will have to delete this thread shortly.

thanks.

[nilendud]

This question is from the new edition of gmat grep..

-Nilendu

[spoorthy]

Top line X X X X
Bottom Line Y Y Y Y Y

To form a triangle you have choose 2 Xs and 1 Y or 2 Ys and 1X.
2Xs can be chosen in 4C2/2! = 4!/2!2! = 6
1 Y can be chosen in 5C1 = 5 ways
So 2Xs and 1 Y can be chosen in 30 ways.

Similarly, 2Ys and 1 X can be chosen in 5!/3!2! * 4 = 40

So total = 30 + 40 = 70.
Lol.. this is exactly what you wrote. All the talk about top and bottom makes it a lil confusing I guess. But your thought process is right.

[RonPurewal]

as in the case of surprisingly many -- probably most -- combinatorics problems, you can also solve this one by just LISTING OUT POSSIBILITIES once you have the basic concept nailed down.

let's say that the points on one line are called a, b, c, d, and the points on the other line are called 1, 2, 3, 4, 5.

case 1: two lettered points and one numbered point:
there are six ways to get two lettered points:
ab, ac, ad, bc, bd, cd
... and there are five ways to get one numbered point (1, 2, 3, 4, or 5).
so, the number of ways to make a line in this way is 6 x 5 = 30.

case 2: two numbered points and one lettered point:
there are ten ways to get two numbered points:
12, 13, 14, 15
23, 24, 25
34, 35
45
... and there are four ways to get one lettered point (a, b, c, or d).
so, the number of ways to make a line in this way is 10 x 4 = 40.

these ways are mutually exclusive -- i.e., no line is in both lists, since they're made in fundamentally different ways -- so you can add these to get 30 + 40 = 70 lines in total.

if you are stuck in combinatorics, IMMEDIATELY start LISTING THINGS.

[lpraat]

Another way:
If there are two parallel lines,one with 4 points and the other with 5 points.

THen:
total number of triangles which can be formed with 9 points: 9C3
=84
Since 4 are on a straight line, we have to neglect the triangles that can be formed from these 4 points= 4c3=4

Similarly neglecting triangles formed from the 5 points of the other line= 5C3=10
therefore, the total number of triangles = 84-4-10=70

[jlucero]

That way works as well. Combination problems are about finding all the ways that work or finding all the ways and subtracting the ways that don't work.