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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:

P_{10} = 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 n_{1} sometimes not_{2} times and not_{no} 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:

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 ___ ___ ___ ___ ___ ___ ___

↓ ↓

1P^{2,3}_{7}

1. P^{2,3}_{7} = __7!__ = 420

2!. 3!

Therefore, with the word BARRIER we can form 420 anagrams.

by Danielle from Miranda

Graduated in Mathematics

**Source:** Brazil School - https://brasilescola.uol.com.br/matematica/permutacao-com-elementos-repetidos.htm