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

Question

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

Bunuel · Accepted Answer

Approach #1:

There are two types of triangles possible: 
With two vertices on the line with 4 points and the third vertex on the line with 5 points -->  C^2_4*C^1_5=30 ;
With two vertices on the line with 5 points and the third vertex on the line with 4 points -->  C^2_5*C^1_4=40 ;

Total:  30+40=70 .

Answer: D.

Approach #2:

All  different  3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear.
 C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70  (where  C^3_4  and  C^3_4  are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).

Answer: D.

Hope it's clear.

BDSunDevil · Answer

say line A with 4 points and line B with 5 points. and triangle is PQR.
If P is on line A and QR on line B we can have 4C1*5C2=40 triangles.
If P is on line B and QR on line A we can have 4C2*5C1=30 triangles.
Total =70 triangles.

mbaiseasy · Answer

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

samsonfred76 · Answer

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

vad3tha · Answer

I really like your 2nd approach :-D

dukenukem · Answer

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

ENGRTOMBA2018 · Answer

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

mcelroytutoring · Answer

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.

ScottTargetTestPrep · Answer

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

Basshead · Answer

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.

Fdambro294 · Answer

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" =   * 5 =   * 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)

ArnauG · Answer

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.

NEPALI_FINOVATE · Answer

I have another way of looking at this question.

Let the line with 4 points be A and 5 points be B for clarifying the solution with you.

So, all together there are 9 points. From these 9 points we can choose 3 points in 9C3 = 84 ways.

Now you cannot have those 3 points in the same line so all 3 points from A OR all 3 points from B is not permissible.

Ways to get all 3 points from A = 4C3 = 4 ways
Ways to get all 3 points from B = 5C3 = 10 ways

So all possible triangles = 84 - (10 +4) = 70 ways.

I like this way better as you can solve triangle problem in rectangular grids this way. Adding all possible from each line can be time consuming for rectangles( 4 or 5 lines cases). 
JUST find out total ways - collinear ways (3 points in same line) and you are set for these triangle type geometry combinatorics problems

bumpbot · Answer

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

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