Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

__Solution:__

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permuatations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

This is a combination problem: combining 2 items out of 3 and is written as follows:

_{n} C _{r} = n! / [ (n - r)! r! ]

The number of combinations is equal to the number of permuations divided by r! to eliminates those counted more than once because the order is not important.

Example 7: Calculate

_{3} C _{2}

_{5} C _{5}

Solution:

_{3} C _{2} = 3! / [ (3 - 2)!2! ] = 6 / [1 * 2] = 3 (problem of points and lines solved above in example 6)

_{5} C _{5} = 5! / [(5 - 5)!5! ] = 5! / [0!5!] = 5! / [1 * 5!] = 1 (there is only one way to select (without order) 5 items from 5 items and to select all of them once!)

Example 8:We need to form a 5 a side team in a class of 12 students. How many different teams can be formed?

Solution:

There is nothing that indicates that the order in which the team members are selected is imoportant and therefore it is a combination problem. Hence the number of teams is given by

_{12} C _{5} = 12! / [ (12 - 5)!5! ] = 792