If 4 points are indicated on a line and 5 points are indicated on

Not that fun this question I wonder how often I would see such a question on test day.

Solution:

* * * * * (line B)

* * * * (line A)

Choose 2 points from line A and a point from line B: = 4!/2!2! * 5!/1!4! = 30
Choose 1 point from line A and 2 points from line B: = 4!/1!3! * 5!/2!3! = 4 * 10 = 40

Combine possiblities: 30 + 40 = 70

See Image below for what my scratch paper looked like on this one

I really like your 2nd approach

Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?

Combination problems are located at search.php?search_id=tag&tag_id=52

Attached is a visual that should help. Bunuel's method is the most elegant, but here is an alternative strategy if the other solution is not clear.

We have two scenarios:

The first scenario is that the 1 vertex is on the first line and the 2 other vertices are on the second line.
So the number of ways to create such a triangle is:

4C1 x 5C2 = 4 x (5 x 4)/2 = 40

The second scenario is that 2 vertices are on the first line and the 1 vertex is on the second line. So the number of ways to create such a triangle is:

4C2 x 5C1 = (4 x 3)/2 x 5 = 6 x 5 = 30

Thus, the total number of ways to create the triangle is 40 + 30 = 70.

Answer: D

1st line: 4 points

2nd line: 5 points

If we choose 1 point on the first line (4C1) and two points from the second line (5C2), we have 4C1 * 5C2 = 40 triangles.

Now, we can choose 2 points on the first line (4C2) and 1 point on the second line (5C1) to get 4C2 * 5C1 = 6 * 5 = 30 triangles

40 + 30 = 70 triangles

Answer is D.

Out of all the Unique Triangles Possible, every single one of them will have 2 Collinear Points on either the:

-Parallel Line with 5 Dots
or
-Parallel Line with 4 Dots

the 3rd Vertex of the Triangle will have to be on the OTHER Parallel Line in order to create a Triangle with positive area


Case 1: We place the Base on the Bottom Parallel Line with 5 Points

(1st)We need to figure out how many Unique Pairs of 2 Points we can pick out of the 5 Points available on this line.

"5 choose 2" = 5!/2!3! = 10 Unique Combinations

AND

(2nd)For each of these 10 unique combinations, there are 4 Points on the Parallel Line on Top which can combine with the given Pair of Points to create a Triangle with positive area. How many ways can we choose 1 for each of the 10 unique combinations:

"4 choose 1" = 4 ways

(10) * (4) = 40 Unique Triangles in which the "Base" is on the Bottom Parallel Line with 5 Dots


Case 2: We place the Base on the Top Parallel Line with 4 Dots.

Same Logic as above, except the Numbers change:

(1st) "4 choose 2" * "5 choose 1" = [4!/2!2!] * 5 = [6] * 5 = 30 Unique Triangles in which the "Base" is on the Top Parallel Line with 4 Dots


40 Triangles from Case 1 + 30 Triangles from Case 2 =

70 Possible Triangles
(D)

To form a triangle, we need three non-collinear points.

Among the 9 given points, we have 4 points on one line and 5 points on another parallel line.

To form a triangle, we can choose one point from the line with 4 points and two points from the line with 5 points, or vice versa.

The number of triangles formed is the combination of choosing 1 point from the line with 4 points and 2 points from the line with 5 points, plus choosing 2 points from the line with 4 points and 1 point from the line with 5 points.

Combining these two cases, we have:

C(4, 1) * C(5, 2) + C(4, 2) * C(5, 1)
= 4 * (5 * 4 / (2 * 1)) + (4 * 3 / (2 * 1)) * 5
= 4 * 10 + 6 * 5
= 40 + 30
= 70.

Therefore, the total number of triangles that can be formed using the 9 points is 70.

Hence, the correct answer is D. 70.