Using the arrangements formula, it is found that they find 46200 ways.
The number of arrangements of n elements is given by:
[tex]A_n = n![/tex]
If they repeat in [tex]n_1, n_2, \cdots, n_n[/tex] ways, the number of arrangements is given by:
[tex]A_n^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2!\cdots n_n}[/tex]
In this problem, 11 elements are arranged with repetitions of 4, 3 and 3, hence:
[tex]T = \frac{11!}{4!3!3!} = 46200[/tex]
They find 46200 ways.
More can be learned about arrangements at https://brainly.com/question/2259733
