Ten distinct noncollinear points are chosen on a plane. a) How many line segments can be formed by joining any two of these points?
A line segment is formed by choosing any two points and joining them
So it comes to C(10,2)
= 10!/(8!,2)
= 45
To find the number of line segments that can be formed by joining any two of these ten distinct noncollinear points, we can use the formula for combinations.
In this case, we want to choose 2 points out of the 10 points. The order in which we choose the points does not matter, and repetition is not allowed since we want distinct points. Therefore, we can use the formula for combinations, which is given by:
nCk = n! / (k!(n-k)!),
where n is the total number of objects to choose from and k is the number of objects we want to choose.
In our case, n = 10 (the number of points), and k = 2 (since we want to choose 2 points to form a line segment).
Now let's substitute these values into the formula:
10C2 = 10! / (2!(10-2)!)
= 10! / (2!8!)
= (10 * 9 * 8!) / (2! * 8!)
= (10 * 9) / 2
= 45.
Therefore, there are 45 line segments that can be formed by joining any two of these ten distinct noncollinear points on a plane.
Note: "Noncollinear" means that no three points lie on the same line.