What is the number of triangles formed by joining 12 points 7 of which are in the same straight line?

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is 185.

Explanation:

Total number of triangles formed from 12 points taking 3 at a time = 12C3

But given that out of 12 points, 7 are collinear

So, these seven points will form no triangle.

∴ The required number of triangles = 12C3 – 7C3

= `(12!)/(3!  9!) - (7!)/(3!4!)`

= `(12 xx 11 xx 10 xx 9!)/(3 xx 2 xx 1 xx 9!) - (7 xx 6 xx 5 xx 4!)/(3 xx 2 xx 1 xx 4!)`

= `(12 xx 11 xx 10)/(3 xx 2) - (7 xx 6 xx 5)/(3 xx 2)`

= 220 – 35

= 185

If 7 points out of 12 are in the same straight line, then the number of triangles formed is(A) 19(B) 158(C) 185(D) 201

Answer

Verified

Hint: First of all, find the ways formed triangles by 12 points that is \[{}^{12}{C_3}\] then after 7 points are in a straight line so find the ways formed triangles by 7 points that is \[{}^7{C_3}\]then subtract this from \[{}^{12}{C_3}\] so, we get the answer.
The formula to calculate \[{}^n{C_r}\] is:
 \[{}^n{C_r} = \left[ {\dfrac{{n!}}{{r!(n - r)!}}} \right]\]

Complete step by step solution:
To formed a triangle, we need to three points so if 12 points are formed \[{}^{12}{C_3}\] triangles.
Total triangle formed by 12 points \[ = {}^{12}{C_3}\]
By using this formula \[{}^n{C_r} = \left[ {\dfrac{{n!}}{{r!(n - r)!}}} \right]\]we can expand \[{}^{12}{C_3}\]like this
\[ = \dfrac{{12!}}{{(3!)(12 - 3)!}}\]
In above equation 12! Expand by using formula \[n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ........1\]
\[ = \dfrac{{12 \times 11 \times 10 \times 9!}}{{(3 \times 2 \times 1)(9!)}}\]
\[ = 220\]triangles
But in a statement that clearly stated as that seven points are in the same line that seven points formed triangle \[{}^7{C_3}\]ways so that seven points not formed a triangle with each other so subtract this way of the Striangle from the total ways of the triangle.
So, If the 7 points out of 12 are \[ = {}^{12}{C_3} - {}^7{C_3}\]
By using this formula \[{}^n{C_r} = \left[ {\dfrac{{n!}}{{r!(n - r)!}}} \right]\]we can expand \[{}^7{C_3}\]like this
\[ = 220 - \left[ {\dfrac{{7!}}{{3!(7 - 3)!}}} \right]\]
\[ = 220 - 35\]
\[ = 185\] triangles

$\therefore$ The number of triangles formed is 185.

Note:
If there is $x$ point to form a triangle the ways of the triangle to form are \[{}^x{C_3}\]. But not to forget if any points like p point are in a straight line then this point form \[{}^p{C_3}\] triangle not to forgot to subtract this triangle.

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Solution

The correct option is A185Explanation for the correct option:Calculate the number of triangles that can be formed.We are given, 12 set of points.Number of triangles formed with 12 points = C312But, according to the given condition, 7 points lie on the same straight lineThus, the selection of 3 points out of 7 collinear points =C37, which we need to deduct from the non-collinear points.Thus, the required number of triangles =C312–C3 7=220-35=185Hence, Option (A) is the correct answer. (adsbygoogle = window.adsbygoogle || []).push({});

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