# Permutations and Combinations Problems

Permutations and combinations are used to solve problems.

## Factorial

Example 1: How many 3 digit numbers can you make using the digits 1, 2 and 3 without repetitions?
method (1) listing all possible numbers using a tree diagram. We can make 6 numbers using 3 digits and without repetitions of the digits.
method (2) counting:
LOOK AT THE TREE DIAGRAM ABOVE.
We have 3 choices for the first digit, 2 choices for the second digit and 1 choice for the third digit.
Using the counting principle, we can say:
The total number of 3-digit numbers is given by
3 � 2 � 1 = 6

There is a special notation for the product 3 � 2 � 1 = 3! and it is read 3 factorial.
In general n! is read n factorial and is given by
n! = n�(n - 1)�(n - 2)�...�2�1

We also define 0! = 1.
Example 2: How many different words can we make using the letters A, B, E and L ?
Solution: We have 4 choices for the first letter, 3 choices for the second letter, 2 choices for the third letter and 1 choice for the fourth letter. Hence the number of words is given by
4 � 3 � 2 � 1 = 4! = 24

## Permutations

Example 3: How many 2 digit numbers can you make using the digits 1, 2, 3 and 4 without repeating the digits?
This time we want to use 2 digits at the time to make 2 digit numbers.
For the first digit we have 4 choices and for the second digit we have 3 choices (4 - 1 used already). Using the counting principle, the number of 2 digit numbers that we can make using 4 digits is given by
4 � 3 = 12

The above problem is that of arranging 2 digits out of 4 in a specific order. This is also called permutating.
The most important idea in permutations is that order is important. When you use the digits 3 and 4 to make a number, the number 34 and 43 are different hence the order of the digits 3 and 4 is important.
In general permutating r (2 digit in the above example) items out of a set of n (4 digits in the above example) items is written as n P r and the formula is given by
n P r = n! / (n - r)!

Example 4: Calculate
4 P 2
6 P 5
4 P 4
Solution:
4 P 2 = 4! / (4 - 2)! = 24/2 = 12
6 P 5 = 6! / (6 - 5)! = 6�5�4�3�2�1 / 1! = 720
4 P 4 = 4! / (4 - 4)! = 4! / 0! = 4! = 4�3�2�1 = 24 (We now understand the need to define 0! = 1)
Example 5: How many 3 letter words can we make with the letters in the word LOVE?
Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. The number of words is given by
4 P 3 = 4! / (4 - 3)! = 24

## Combinations

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 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:

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

## Problems

1. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions?
2. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions?
3. How many 6 letter words can we make using the letters in the word LIBERTY without repetitions?
4. In how many ways can you arrange 5 different books on a shelf?
5. In how many ways can you select a committee of 3 students out of 10 students?
6. How many triangles can you make using 6 non collinear points on a plane?
7. A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. How many different committee can be formed from the group?
8. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. How many number plates can be formed if neither the digits nor the letters are repeated?

## solutions to the above problems

1. 4! = 24
2. 5 P 3 = 60
3. 7 P 6 = 5040
4. 5! = 120
5. 10 C 3 = 120
6. 6 C 3 = 20
7. 10 C 312 c 4 = 59,400
8. 9 P 426 P 3 = 47,174,400

### More References and links

elementary statistics and probabilities. Combinations Calculator. Calculate the number of combinations of n elements taken r at the time.
Permutations Calculator. Calculate the number of permutations of n elements taken r at the time.