Lesson # 58: Permutations Involving Identical Objects

Objective: to find the number of permutation of n object when some of the objects are identical.

1) The permutations of the 4 different letter A, B, E and F are:

4! = 24

2) What happens if two of the letter are the same? Investigate this by converting each F to and E in the list below. Then count the number of permutations of the letters A, B, B and E.

4! = 12

2!

3) What happens if three of the letter are the same? Investigate this by converting each F and E to a B in the list below. Then count the number of permutations of the letters A, B, B and E.

4! = 4

3!

4a) A, B, C, D, D = 5! = 60
                                2!
b) A, B, B, D, D =   5!  = 30
                             2!x2!
Generalization:  The number of permutations of n objects of which there are a objects alike of one kind, b alike of another kind, c alike of another kind, and so one, is...      n!   
                                                                                              a! b! c!

        8!          

2!x2!x2!x2!

Lesson # 57: Permutations Involving Different Objects

Objectives: to introduce the concept of permutation
                  to find the number of permutations of n district objects taken r at a time

example: Two letter, A and B, can be written in two different orders, AB and BA. These are permutations of A and B.
a)  List all of the permutations of 3 letter A, B and C.
ABC, ACB, BAC, BCA, CAB, CBA
                 3 x 2 x 1 = 6
b) How many permutations are there for 4 letters A, B, C and D.
             4 x 3 x 2 x 1 = 24
c) Predict the number of permutations of 5 letters A, B, C, D and E
           5 x 4 x 3 x 2 x 1 = 120        5! (factorial)

example: Consider the letters A, B, C, D and E.  Instead of using all the letters to from permutations, we could use fewer letters.  For example, DB is a 2-letter perutations of these 5 letters.
a) List all the different 2-letter permutation of the 5 letter A, B, C, D and E.
AB AC AD AE     DA DB DC DE                        cP2
BA BC BD BE      EA EB EC ED    = 20              5P2 = 20
CA CB CD CE            

5 x 4 = 20    5P2 =    5!       = 5 x 4 x 3 x 2 x 1      EQUATION:  nPr =  n!
                               (5-2)!                  3 x 2 x 1                                       (n-r)!

example: How many different permutation can be formed using all of the letter of the word COMPUTE?
7P7  or   7! = 5040