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.

tree diagram for numbers made from 3 digits


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 permuatations 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 formuala 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 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

Problems with solutions

  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
  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 3 * 12 c 4 = 59,400

  8. 9 P 4 * 26 P 3 = 47,174,400




More references on
elementary statistics and probabilities.


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Updated: 2 April 2013

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