Permutation of repeated elements must follow a different form from permutation, as repeated elements interchange with each other. To understand how this happens, see the example below:
The permutation of the word MATHEMATICS would look like this:
Without taking into account the repeated letters (elements), the permutation would look like this:
P10 = 10! = 3.628.800
Now, as the word MATHEMATICS has elements that repeat, like the letter A that repeats 3 times, the letter T repeats 2 times and letter M repeats 2 times, so the permutation between each other of these repetitions would be 3!. 2!. 2!. Therefore, the permutation of the word MATHEMATICS will be:
Therefore, with the word MATHEMATICS we can assemble 151200 anagrams.
Following this reasoning, we can conclude that, in general, the permutation with repeated elements is calculated using the following formula:
Given the permutation of a set with n elements, some elements repeat n1 sometimes not2 times and notno times. Then the permutation is calculated:
Example 1:
How many anagrams can be formed with the word MARAJOARA, applying the permutation we will have:
Therefore, with the word MARAJOARA we can form 7560 anagrams.
Example 2:
How many anagrams can be formed with the word ITALIAN, applying the permutation we will have:
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So with the word ITALIAN we can form 3360 anagrams.
Example 3:
How many anagrams with the word BARRIER can be formed, which must start with the letter B?
B ___ ___ ___ ___ ___ ___ ___
↓ ↓
1P2,37
1. P2,37 = 7! = 420
2!. 3!
Therefore, with the word BARRIER we can form 420 anagrams.
by Danielle from Miranda
Graduated in Mathematics
Would you like to reference this text in a school or academic work? Look:
RAMOS, Danielle de Miranda. "Permutation with repeated elements"; Brazil School. Available in: https://brasilescola.uol.com.br/matematica/permutacao-com-elementos-repetidos.htm. Accessed on June 28, 2021.
