If A, B, C, D, E and F are 6 distinct points on the circumference of a circle, how many different chords can be drawn using any 2 of the 6 points ?

Options

A. 6
B. 12
C. 15
D. 30

Explanation

To find the number of different chords that can be drawn using any 2 of the 6 distinct points (A, B, C, D, E, and F) on the circumference of a circle, we can use the concept of combinations.

### Step-by-Step Solution:

1. **Identify the Total Points**: We have 6 distinct points on the circumference of the circle. Let's denote them as A, B, C, D, E, and F.

2. **Determine the Combination Formula**: To find the number of ways to choose 2 points from these 6 points, we use the combination formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of points, and \( r \) is the number of points to choose.

3. **Substitute Values**: Here, \( n = 6 \) (the total number of points) and \( r = 2 \) (the number of points we want to choose to form a chord). Thus, we need to calculate \( 6C2 \):
\[ 6C2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} \]

4. **Simplify Factorials**: We can simplify \( 6! \) as follows:
\[ 6! = 6 \times 5 \times 4! \]
Therefore, we can rewrite the combination as:
\[ 6C2 = \frac{6 \times 5 \times 4!}{2! \times 4!} \]

5. **Cancel Out the Factorials**: The \( 4! \) in the numerator and denominator cancels out:
\[ 6C2 = \frac{6 \times 5}{2!} \]

6. **Calculate \( 2! \)**: We know that \( 2! = 2 \times 1 = 2 \). Now substitute this value:
\[ 6C2 = \frac{6 \times 5}{2} = \frac{30}{2} = 15 \]

7. **Conclusion**: Therefore, the number of different chords that can be drawn using any 2 of the 6 points is:
\[ \text{Number of chords} = 15 \]