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Combinatorics - Probability
Permutations with identical elements

Identical elements
Permutations

Permutations with identical elements


Let's consider another example:
How many ways do we have to arrage (permute) 7 books when 3 are the same?
B1, B2, B2, B3, B4, B2, B5


Let's disguish the three same books B2, by writing the set: B1, B21, B22, B3, B4, B23, B5
Here the number of permutations (7 arrangement among 7) is 7!

For the set {B1, B2, B2, B3, B4, B2, B5}, we have 3! permutations of B2 that not affect the place of the four remaing elements B1, B3, B4 and B5.

Hence, for each permutation of 7 elements that we have for the set {B1, B2, B2, B3, B4, B2, B5}, 3! permutations are identical.

Then:
The number of permutations of 7 different elements is equal to (the number of permutations of 7 elements wich contains 3 identical elemnts) x (the number of permutations of the 3 identical elements), that is:
(the number of permutations of 7 elements wich contains 3 identical elemnt) x 3! = 7!

In the general case, If a set contains n elements whith m1 identical elements of a certain kind, m2 identical elements of another kind, m3 identical elements of another kind, ..., and mk identical elements of another kind, the number of permutations is equal to
n!/m1 x m2 x m3 x ... x mk = n!/?mi (i from 1 to k)


P(mi , n) = n!/?mi
(i from 1 to k)


Examples:

1. The number of ways to write the word "Goooogle" (8 letters in wich 4 are identical) is:
8!/!4 = 5 x 6 x 7 x 8 = 1680 possiblities.

2. How many words can we form with the letters S, E, A, R, C, H; but the two letters E and A must be neighbors?

Here, we will permute three letters S, R, C, H and the group {E, A}. There are 5! possibilities (as if we permute 5 letters). For each of these possibilities, we have 2! possibilities to permute the letters of the group {E, A}.
Then, in total, we have 5! x 2! = 240.

We have then 240 ways to write a word from the letters: S, E, A, R, C, H; but the two vowels E and A will remain together.





  
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