Six points are taken on a circle. The number of triangles formed inside the circle is
 $ \left( a \right){\text{ 20}} $
 $ \left( b \right){\text{ 22}} $
 $ \left( c \right){\text{ 25}} $
 $ \left( d \right){\text{ 32}} $

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Hint: Here in this question it is given that there are a total of six points on a circle. And as we know when the points are on a circle so when joining any three points will form a triangle. Therefore the number of ways will be calculated as the number of ways of selecting three points out of six. So by using a combination, we will solve this problem.
Formula used:
The combination is given by,
 $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Here, $ n $ will be the number of items in the set and
 $ r $ , will be the number of items selected from the set.

Complete step-by-step answer:
 Here in this question we have a number of points on the circle which will be equal to $ 6 $ and will be denoted by $ n $ .
 $ r $ will be the number of points required to define a triangle and so it will be equal to $ 3 $ . So we will now find the combination of six items taken three at a time.
So mathematically it can be written as
 $ \Rightarrow {}^6{C_3} $
Since the formula for solving it is given by $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
So on substituting the values, we get the RHS equals to
 $ \Rightarrow \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} $
On solving the braces of the denominator, we get the above equation as
 $ \Rightarrow \dfrac{{6!}}{{3! \times 3!}} $
On expanding the numerator and denominator, we get
 $ \Rightarrow \dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2 \times 1 \times 3!}} $
On canceling the like term and reducing the fraction into the simplest form possible, we get
 $ \Rightarrow 20 $
Therefore, the number of triangles formed inside the circle is $ 20 $ .
Hence, the option $ \left( a \right) $ is correct.
So, the correct answer is “Option a”.

Note: Since there is a close concept related to the permutation and permutation will take order into account. So if the questions have asked about a specific use case where all the triangles would be considered as different then in that case we have needed to use permutations.