Seven distinct points are identified on the circumference of a circle. How many different triangles can be formed if each vertex must be one of these 7 points?
To form a triangle, we must choose 3 of the 7 points to be the vertices. So, selecting 3 points (without concern for order) out of 7 we get ${7 \choose 3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = \boxed{35}$ triangles.
35
Level 2
Counting & Probability
Problem: Seven distinct points are identified on the circumference of a circle. How many different triangles can be formed if each vertex must be one of these 7 points?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of different triangles that can be formed from seven distinct points on a circle, we need to understand the properties of combinations in combinatorial mathematics. A triangle is defined by three vertices, and since the order of selection does not matter, we can use the combination formula.

The combination formula is given by:

C(n, k) = n! / (k! * (n - k)!)

In this scenario, n represents the total number of points available, which is 7, and k represents the number of points needed to form a triangle, which is 3. Therefore, we need to calculate C(7, 3).

Substituting the values into the formula:

C(7, 3) = 7! / (3! * (7 - 3)!)
C(7, 3) = 7! / (3! * 4!)

Next, we calculate the factorials:

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
3! = 3 × 2 × 1
4! = 4 × 3 × 2 × 1

Now, we can simplify C(7, 3):

C(7, 3) = (7 × 6 × 5) / (3 × 2 × 1)
C(7, 3) = 210 / 6
C(7, 3) = 35

Thus, the total number of different triangles that can be formed using the seven distinct points on the circumference of the circle is 35.<|END_OF_LATENT|> The final answer is $35$.