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