Permutation with repeated elements

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 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^2,3₇
1. P^2,3₇ = 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