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Solution


Given:
    • Total number of points = 12
      o Number of collinear points = 7

To find:
    • The number of triangles that can be formed

Approach and Working:
    • The number of triangles that can be formed, N = number of ways of selecting 2 points from 7 and the other from the remaining 5 + the number of ways of selecting 1 from 7 and 2 from the remaining 5 + the number of ways of selecting 3 points from the remaining 5
    • Therefore, \(N = ^7C_2 * ^5C_1 + ^7C_1 * ^5C_2 + ^5C_3 = 21 * 5 + 7 * 10 + 10 = 185\)

Hence, the correct answer is Option C

Answer: C

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Bunuel
How many triangles can be formed by joining 12 points, 7 of which are collinear?

A. 255
B. 220
C. 185
D. 35
E. 10

We can start with 12C3 to get \(\frac{12*11*10}{3!} = 2*11*10\) random selections.

The problem with that random selection is the 7 collinear points, if we pick 3 points on the same line it will not form a triangle so we need to eliminate 7C3 = \(\frac{7*6*5}{3!} = 7*5\) selections.

2*11*10 - 7*5 = 220 - 35 = 185.

Ans: C
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We can solve by finding total triangles formed by 12 points and then subtract those by the total triangles formed by 7 points ( as they are collinear ).

12C3-7C3= 185. Hence,C

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Total triangles that can be formed using the \(12\) points are \(12C3 = 220\)
But, \(7\) of these points are collinear, and hence can't be used to form triangles among themselves.

So, number of triangles that can't be formed are \(7C3 = 35\)

\(220-35 = 185\)

Answer C.
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Bunuel
How many triangles can be formed by joining 12 points, 7 of which are collinear?

A. 255
B. 220
C. 185
D. 35
E. 10
We can draw a line through the 7 points since they are collinear. Using this information, there are three cases:

→ Case 1: Choose two collinear points and another which isn't collinear = \(7C_2\)*\(5C_1\) = 105
→ Case 2: Choose one collinear point and two points which aren't collinear = \(7C_1\)*\(5C_2\) = 80
→ Case 3: Combine three points which aren't collinear = \(5C_3\) = 10
→ Total cases = 105+80+10 = 185.
→ Option C is the correct answer choice.
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