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The problem reads:

In a classroom there are $16$ students, $4$ rows of desks, each made up of $4$ desks.

  1. How many different ways are there to distribute the students over the desks?
  2. Assuming there are $4$ female students, and $12$ male students, how many ways are there to distribute the students over the desks so that all four females aren't in the same row?

Point $(1)$ was pretty straightforward in my opinion: we are trying to find the number of functions: $$f : \{1,2,...,16\} \rightarrow \{1,2,...,16\}$$ that are injective. Since the domain and codomain have the same cardinality, such number is $16!$, that is the number of permutations of a set of $16$ elements.

For point $(2)$, we have to take the number of permutations from the previous point, and subtract the permutations where all four females are sitting in the same row. Since there are $4$ females, the number of arrangements of those four students over a row of desks is $4!$. Since there are $4$ rows, the configurations that have $4$ female students, in any order, all sitting in the same one row out of the four, is $4\cdot4!$.

Therefore, the final answer is $16! - 4\cdot4!$. This is the first combinatory problem I try to solve so I'm not sure if my reasoning is correct. Feedback is much appreciated.

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    $\begingroup$ You were close. It should be $16!-4{\,\cdot\,}4!{\,\cdot\,}12!$.$\;$Do you see why we need the factor $12!$? $\endgroup$ – quasi Apr 12 at 15:56
  • $\begingroup$ Is that because, for every one of the $4\cdot4!$ cases, the remaining $12$ students can be arranged in $12!$ ways? $\endgroup$ – Samuele B. Apr 12 at 15:58
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    $\begingroup$ Yes, exactly.${}{}{}$ $\endgroup$ – quasi Apr 12 at 15:59
  • $\begingroup$ Feel free to post an answer to your own question. $\endgroup$ – quasi Apr 12 at 16:18
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As user quasi made me notice in the comments, the correct answer is: $$16!-4\cdot4!\cdot12!$$ Because, for every one of those $4\cdot4!$ arrangements of the 4 female students over the 4 sets of 4 seats, the remaining $12$ male students can be arranged in $12!$ ways. Thank you for pointing it out to me.

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