Ten points are marked on one side of △ABC. Another side of △ABC has 11 marked points and the third side has 12 marked points. None of the points are a vertex of the triangle. If any three non-collinear points can be the vertices of a new triangle, how many such triangles are possible?
Well, let's see. If we have 10 points on one side, 11 points on another side, and 12 points on the third side, then the total number of points is 10 + 11 + 12 = 33. Since none of these points are vertices of the triangle, we have 33 - 3 = 30 possible points to choose from for each vertex.
Now, to form a triangle, we need to choose 3 points from these 30 possible points. The number of ways to choose 3 points from a set of 30 is given by the combination formula, which is 30 choose 3.
Now, allow me to calculate this for you... drumroll, please...
(30 choose 3) = (30!)/(3!(30-3)!) = 4060.
So, it looks like there are 4060 possible triangles that can be formed using these points. Remember, though, quality over quantity. It's not just about the number of triangles, but also their shape and properties that make them interesting. Happy triangle counting!
To solve this problem, we can use the concept of combinations and the formula for calculating the number of combinations.
Step 1: Calculate the number of ways to choose 3 points from the first side.
Using the formula for combinations, we can calculate the number of ways to choose 3 points from the 10 marked points on the first side of △ABC:
C(10, 3) = 10! / (3! * (10-3)!) = 120.
Step 2: Calculate the number of ways to choose 3 points from the second side.
Similarly, we can calculate the number of ways to choose 3 points from the 11 marked points on the second side of △ABC:
C(11, 3) = 11! / (3! * (11-3)!) = 165.
Step 3: Calculate the number of ways to choose 3 points from the third side.
Finally, we calculate the number of ways to choose 3 points from the 12 marked points on the third side of △ABC:
C(12, 3) = 12! / (3! * (12-3)!) = 220.
Step 4: Calculate the total number of triangles.
Since any three non-collinear points can be the vertices of a new triangle, we can add up the number of triangles from each side:
Total number of triangles = C(10, 3) + C(11, 3) + C(12, 3) = 120 + 165 + 220 = 505.
Therefore, there are 505 possible triangles that can be formed using the marked points on △ABC.
To find the number of possible triangles, we can use the combination formula. Since we need to choose 3 points from a set, we can use the formula for combinations:
nCr = n! / (r!(n-r)!)
Where n is the total number of points on a side of the triangle, and r is the number of points we want to choose (in this case, 3).
For the first side with 10 points, we can choose 3 points out of 10:
nCr = 10! / (3!(10-3)!) = 10! / (3!7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120 / 6 = 20
For the second side with 11 points, we can choose 3 points out of 11:
nCr = 11! / (3!(11-3)!) = 11! / (3!8!) = (11 * 10 * 9) / (3 * 2 * 1) = 990 / 6 = 165
For the third side with 12 points, we can choose 3 points out of 12:
nCr = 12! / (3!(12-3)!) = 12! / (3!9!) = (12 * 11 * 10) / (3 * 2 * 1) = 1320 / 6 = 220
Now, to find the total number of triangles possible, we sum up the number of triangles from each side:
Total number of triangles = 20 + 165 + 220 = 405
Therefore, there are 405 possible triangles that can be formed using the given set of marked points.