Question

If 7 points out of 12 are in the same straight line, then the number of triangles formed is
    (a) 19
    (b) 185
    (c) 201
    (d) None of these

Answer 1

Hint: Here, we have to apply the formula for combination,${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. 3 points are required for a triangle but not all 3 in the same line. To get the number of triangles we have to subtract ${}^{7}{{C}_{3}}$ ways from the total possible ${}^{12}{{C}_{3}}$ ways.

Complete step-by-step solution -
Here, the total number of points is given as 12.
It is also given that 7 points are in the same straight line.
Now, we have to find the total number of triangles formed from the given conditions.
We know that a triangle can be formed by joining 3 points but not all 3 in the same line.
So from the total of 12 points, 3 points can be selected in ${}^{12}{{C}_{3}}$ ways.
But in the question we have 7 points in the straight line, so these 7 points can’t be joined together to form a triangle.
i.e. the number of triangles formed by 7 points are ${}^{7}{{C}_{3}}$.
But we have to avoid the ${}^{7}{{C}_{3}}$possible ways from ${}^{12}{{C}_{3}}$ total possible ways, since 7 points are in the straight line.
Therefore, the total number of triangles formed = ${}^{12}{{C}_{3}}-{}^{7}{{C}_{3}}$
We know by combinations that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Therefore by the above formula, we can write:
$\begin{align}
  & {}^{12}{{C}_{3}}=\dfrac{12!}{3!(12-3)!} \\
 & {}^{12}{{C}_{3}}=\dfrac{12!}{3!\text{ }9!}\text{ }.....\text{ (1)} \\
\end{align}$
We know that,
 $\begin{align}
  & 12!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12 \\
 & 9!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9 \\
 & 3!=1\times 2\times 3 \\
\end{align}$
Now, by substituting all these values in equation (1) we get,
${}^{12}{{C}_{3}}=\dfrac{1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12}{1\times 2\times 3\times 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9}$
Next, by cancellation we obtain:
$\begin{align}
  & {}^{12}{{C}_{3}}=10\times 11\times 2 \\
 & {}^{12}{{C}_{3}}=220\text{ }.....\text{ (2)} \\
\end{align}$
Similarly, we can write:
$\begin{align}
  & {}^{7}{{C}_{3}}=\dfrac{7!}{3!(7-3)!} \\
 & {}^{7}{{C}_{3}}=\dfrac{7!}{3!\text{ 4}!}\text{ }.....\text{ (3)} \\
\end{align}$
We know that,
$\begin{align}
  & 7!=1\times 2\times 3\times 4\times 5\times 6\times 7 \\
 & 4!=1\times 2\times 3\times 4 \\
 & 3!=1\times 2\times 3 \\
\end{align}$
Now, by substituting all these values in equation (3) we get,
${}^{7}{{C}_{3}}=\dfrac{1\times 2\times 3\times 4\times 5\times 6\times 7}{1\times 2\times 3\times 1\times 2\times 3\times 4}$
Next, by cancellation we obtain:
$\begin{align}
  & {}^{7}{{C}_{3}}=5\times 7 \\
 & {}^{7}{{C}_{3}}=35\text{ }.....\text{ (4)} \\
\end{align}$
From equation (2) and equation (3) we can write:
$\begin{align}
  & {}^{12}{{C}_{3}}-{}^{7}{{C}_{3}}-220-35 \\
 & {}^{12}{{C}_{3}}-{}^{7}{{C}_{3}}=185 \\
\end{align}$
Therefore, the number of triangles can be formed = $185$
Hence, the correct answer for this question is option (b)

Note: Here out of 12 points 7 are in the straight line, therefore, it cannot be joined to form a triangle. i.e. while finding the number of triangles we have to subtract ${}^{7}{{C}_{3}}$ ways from total ${}^{12}{{C}_{3}}$ ways. Otherwise you will get a wrong answer.