Out of 15 points in a plane, n points are in the same straight line. 4...
Solution:
To solve this problem, we need to use the concept of combinations and counting principles.
Step 1: Counting the number of triangles formed by selecting any 3 points out of 15.
The number of ways to select 3 points out of 15 is given by the combination formula C(15, 3):
C(15, 3) = 15! / (3!(15-3)!) = 15! / (3!12!) = (15 * 14 * 13) / (3 * 2 * 1) = 455
So, there are a total of 455 triangles that can be formed by selecting any 3 points out of the 15.
Step 2: Counting the number of triangles formed by selecting 3 points in a straight line.
If n points are in the same straight line, then any triangle formed by selecting 3 points from these n points will be a straight line itself and hence not a valid triangle. So, we need to subtract the number of straight lines from the total number of triangles.
The number of straight lines that can be formed by selecting any 2 points out of n is given by the combination formula C(n, 2):
C(n, 2) = n! / (2!(n-2)!) = n! / (2! (n-2)!) = (n * (n-1)) / (2 * 1) = (n^2 - n) / 2
So, the number of triangles formed by selecting 3 points in a straight line is given by:
Number of triangles = C(n, 3) - C(n, 2) = (n! / (3!(n-3)!)) - ((n^2 - n) / 2)
We are given that the number of triangles formed by selecting any 3 points out of 15 is 445. So, we can write the equation as:
445 = C(15, 3) - ((n^2 - n) / 2)
Substituting the value of C(15, 3) = 455 in the equation, we get:
445 = 455 - ((n^2 - n) / 2)
Simplifying the equation, we have:
10 = (n^2 - n) / 2
Multiplying both sides of the equation by 2, we get:
20 = n^2 - n
Rearranging the equation, we have:
n^2 - n - 20 = 0
Factoring the quadratic equation, we have:
(n - 5)(n + 4) = 0
So, n = 5 or n = -4