distint points a, b, c, d, e, and f are marked on a circle. How many triangles can be formed using these points as vertices?
Draw a circle, form as many triangles as you can, and count.
ok i tried that that its hard why don't u try and tellme the answer ?
You have 6 points on a circle.
It is not possible to have 3 of them forming a straight line (not a triangle) so the question becomes how many choices are there of 3 elements form 6
or C(6,3) = 6!/(3!3!) = 20
To solve this problem, we can use the concept of combinations or binomial coefficients.
In order to form a triangle, we need to choose 3 points out of the given 6 points (a, b, c, d, e, and f) that are marked on the circle.
To calculate the number of different combinations, we can use the binomial coefficient formula, known as "n choose k," which is represented as (n C k) or nCk.
The formula for binomial coefficient is:
(n C k) = n! / (k! * (n - k)!)
Where n! is the factorial of n.
In our case, n = 6 (the total number of points) and k = 3 (the number of points required to form a triangle).
Using the formula, we can calculate:
(6 C 3) = 6! / (3! * (6 - 3)!)
= 6! / (3! * 3!)
= (6 * 5 * 4 * 3!)/(3! * 3 * 2 * 1)
= (6 * 5 * 4) / (3 * 2 * 1)
= 20
Therefore, there are 20 different triangles that can be formed using these points as vertices.