Combinatorics - Combinatorial Analysis
Variations
Variations or permuted combinations (permutations without repetition)
Variations with repetition (or permuted combinations with repetition)
Variations or permuted combinations (permutations without repetition)
The variations of size r chosen from a set of n different objects are the permutations of combinations of r.
The number of variations of size r chosen from n objects equals the number of combinations of size r multiplied by the r! permutations,
Example:  Find the number of variations of size 3 that can be made from digits 1, 2, 3, 4 and write them out.
Solution:  Since, n = 4 and r = 3 then
Notice that there are 4 combinations of size 3 chosen from the given 4 digits, each of which gives six permutations as is shown below.
The variations are,          1 2 3           1 2 4           1 3 4           2 3 4
1 3 2           1 4 2           1 4 3           2 4 3
2 1 3           2 1 4           3 1 4           3 2 4
2 3 1           2 4 1           3 4 1           3 4 2
3 1 2           4 1 2           4 1 3           4 2 3
3 2 1           4 2 1           4 3 1           4 3 2.
Variations with repetition (or permuted combinations with repetition)
The number of ways to choose r objects from a set of n different objects when order is important and one object can be chosen more than once
Example:  Find the number of variations with repetition of size 4 that can be made from digits  0, 1, 2 and write them out.
Solution:  Since, n = 3 and r = 4 then the total number of the variations with repetition is
As in the above example, first we should select all combinations with repetition of size 4 from the 3 given digits, and then write the permutations of each of them.
Let calculate the number of combinations
The 15 distinct combinations are divided into four groups depending on the number of permutations each group yields,
 a b c d 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 2 1 1 1 1 0 0 0 2 0 0 2 2 1 1 0 2 2 2 2 2 1 1 1 0 1 1 2 2 2 2 0 1 1 1 1 2 2 2 2 0 2 2 2 1
The 3 combinations from the group a can not be rearranged.
Each combination from the group b, c and d, gives,
 b) c) d)
as they are the permutations of the 4 objects some of which are the same.
Thus, the permuted combinations (or the variations) of the group b are,
0 0 0 1     0 0 0 2     1 1 1 0     1 1 1 2     2 2 2 0     2 2 2 1
0 0 1 0     0 0 2 0     1 1 0 1     1 1 2 1     2 2 0 2     2 2 1 2
0 1 0 0     0 2 0 0     1 0 1 1     1 2 1 1     2 0 2 2     2 1 2 2
1 0 0 0     2 0 0 0     0 1 1 1     2 1 1 1     0 2 2 2     1 2 2 2   ( 6 ´ 4 = 24 variations)
the variations of the group c are,                                            the variations of the group d are,
0 0 1 1     0 0 2 2     1 1 2 2                                                  0 0 1 2     1 1 0 2     2 2 0 1
0 1 0 1     0 2 0 2     1 2 1 2                                                  0 0 2 1     1 1 2 0     2 2 1 0
0 1 1 0     0 2 2 0     1 2 2 1                                                  0 1 0 2     1 0 1 2     2 0 2 1
1 1 0 0     2 2 0 0     2 2 1 1                                                  0 1 2 0     1 0 2 1     2 0 1 2
1 0 1 0     2 0 2 0     2 1 2 1                                                  0 2 0 1     1 2 1 0     2 1 2 0
1 0 0 1     2 0 0 2     2 1 1 2                                                  0 2 1 0     1 2 0 1     2 1 0 2
( 3 ´ 6 = 18 variations)                                                      1 0 0 2     0 1 1 2     0 2 2 1
1 0 2 0     0 1 2 1     0 2 1 2
1 2 0 0     0 2 1 1     0 1 2 2
2 0 0 1     2 1 1 0     1 2 2 0
2 0 1 0     2 1 0 1     1 2 0 2
2 1 0 0     2 0 1 1     1 0 2 2
( 3 ´ 12 = 36 variations)
Therefore, the total number of the variations of size 4 with repetition chosen from the given 3 digits are,
V(3, 4) = 3 + 6 ´ 4 + 3 ´ 6 + 3 ´ 12 = 3 · (1 + 8 + 6 + 12) = 3 · 27 = 3 · 33 = 34 = 81.
Example: Find the number of variations with repetition of size 8 that can be made from the binary digits 0, 1.
Solution:  Since, n = 2 and r = 8 then the total number of the variations with repetition is
V(n, r) = nr       =>       V(2, 8) = 28 = 256.
First we should select the combinations of size 8 that can be made from the 2 binary digits, then examine the number of ways each combination can be rearranged or permuted to prove the total number of variations.
So, the number of the combinations is
The 9 combinations are,
1)
 0 0 0 0 0 0 0 0
this combination can not be rearranged or permuted
2)
 1 1 1 1 1 1 1 1
this combination can not be rearranged or permuted
3)
 0 0 0 0 0 0 0 1
4)
 0 0 0 0 0 0 1 1
5)
 0 0 0 0 0 1 1 1
6)
 0 0 0 0 1 1 1 1
7)
 0 0 0 1 1 1 1 1
8)
 0 0 1 1 1 1 1 1
9)
 0 1 1 1 1 1 1 1
Therefore, the total number of the variations of size 8 with repetition chosen from the given 2 digits are,
V(2, 8) = 2 + 2 ´ 8 + 2 ´ 28 + 2 ´ 56 + 70 = 2 · (1 + 8 + 28 + 56 + 35) = 2 · 128 = 2 · 27 = 28
V(2, 8) = 28 = 256.
Both numerical and nonnumerical data can be processed by the computer as all letters digits and special characters are coded (represented as a unique sequence of binary digits 0 and 1) using binary variations.
Intermediate algebra contents