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?
n C r = n! / [ (n - r)! r! ]
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 permutations, 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:
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
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?
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