To determine how many distinct lines are formed by 10 points in a plane where no three points are collinear, we need to calculate the number of ways we can choose pairs of points from those 10. This is a combinatorial problem where we are interested in combinations, not permutations, because the order in which we select the points does not matter.
To calculate the number of distinct lines determined by 10 points taken 2 at a time, we need to use the combination formula, which is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
Where:
- [tex]\( n \)[/tex] is the total number of points,
- [tex]\( r \)[/tex] is the number of points in each pair,
- [tex]\( n! \)[/tex] denotes the factorial of [tex]\( n \)[/tex], which is the product of all positive integers up to and including [tex]\( n \)[/tex].
For our problem:
- [tex]\( n = 10 \)[/tex] (the total number of points),
- [tex]\( r = 2 \)[/tex] (we are choosing pairs of points).
Using the combination formula:
[tex]\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} \][/tex]
[tex]\[ \binom{10}{2} = \frac{10!}{2! \cdot 8!} \][/tex]
We know that [tex]\( 10! \)[/tex] can be expanded as [tex]\( 10 \times 9 \times 8! \)[/tex], so:
[tex]\[ \binom{10}{2} = \frac{10 \times 9 \times 8!}{2! \cdot 8!} \][/tex]
The [tex]\( 8! \)[/tex] terms cancel out:
[tex]\[ \binom{10}{2} = \frac{10 \times 9}{2!} \][/tex]
Now, [tex]\( 2! \)[/tex] (which means 2 factorial) is:
[tex]\[ 2! = 2 \times 1 = 2 \][/tex]
So, the equation simplifies to:
[tex]\[ \binom{10}{2} = \frac{10 \times 9}{2} \][/tex]
[tex]\[ \binom{10}{2} = \frac{90}{2} \][/tex]
[tex]\[ \binom{10}{2} = 45 \][/tex]
Therefore, the number of distinct lines determined by these 10 points taken 2 at a time is [tex]\( 45 \)[/tex].
