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

Last updated date: 11th Sep 2024
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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.

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.