Combinatorics - Combinatorial Analysis Permutations Permutations of n objects some of which are the same
Permutations
Given a set of n different elements or objects. Any distinct ordered arrangement of the n elements is called permutation.
The total number of permutations for n elements is
 P(n) = n!.
Example:   Given is the sequence of four digits 1, 2, 3, 4. Write all possible ordered arrangements or permutations of the 4 digits.
Solution:  The number of permutations of the given 4 digits,  P(4) = 4! = 4 · 3 · 2 · 1 = 24.
The permutations are,
 1, 2, 3, 4         2, 1, 3, 4         3, 1, 2, 4         4, 1, 2, 3 1, 2, 4, 3         2, 1, 4, 3         3, 1, 4, 2         4, 1, 3, 2 1, 3, 2, 4         2, 3, 1, 4         3, 2, 1, 4         4, 2, 1, 3 1, 3, 4, 2         2, 3, 4, 1         3, 2, 4, 1         4, 2, 3, 1 1, 4, 2, 3         2, 4, 1, 3         3, 4, 1, 2         4, 3, 1, 2 1, 4, 3, 2         2, 4, 3, 1         3, 4, 2, 1         4, 3, 2, 1.
Permutations of n objects some of which are the same
The number of permutations of n elements some groups of which are the same where, k1, k2, . . . , km denotes each group with identical elements.
Example:   How many different 7-letter words can be formed from the word GREETER?
 Solution: since the letter R repeats twice and E repeats 3 times.
Example:   How many four-digit numbers can be written with all of the digits 2, 3, 3, 4 and write them in increasing order.
Solution:  In the given sequence of four digits, the digit 3 repeat twice, so the 12 four-digit numbers written in increasing order are,
2 3 3 4         3 2 3 4         4 2 3 3
2 3 4 3         3 2 4 3         4 3 2 3
2 4 3 3         3 3 2 4         4 3 3 2.
3 3 4 2
3 4 2 3
3 4 3 2   Intermediate algebra contents 