# Permutation: 16 students in 2 rows

There are $$8$$ male and $$8$$ female students. We arrange them to sit in $$2$$ separate rows with $$8$$ students in each row. How many ways are there to arrange them if no male can sit next to each other?

My attempt is to arrange the $$16$$ students in one row first, then we call the first $$8$$ students as the one row and the remaining $$8$$ the second row. To have no male sitting next to each other, we first permute the $$8$$ female, we have $$8!$$ arrangements. Then, arrange the males into the 'gaps' in-between the females. There are $$9$$ gaps. Hence, the arrangements for males would be $$9P8$$. The required number of arrangement is $$8! \times 9P8$$.

I would appreciate it if you can point out my mistake and give me some hints about the correct method. Thanks in advance.

• In the row of 16 two males can sit next to each other on places 8,9, as they correspond to different "real" row.
– YJT
Apr 27, 2021 at 8:09

Under the interpretation that no two male students in the same row can be adjacent:

There can be at most four male students in each row since otherwise two male students would have to be adjacent. Since there are a total of eight male students, there must be four male students in each row. Consequently, there are also four female students in each row.

Choose which four female students occupy the front row, which can be done in $$\binom{8}{4}$$ ways. Arrange them in the front row, which can be done in $$4!$$ ways. This creates five spaces in which to place the four males. $$\square f_1 \square f_2 \square f_3 \square f_4 \square$$ To ensure that no two males are adjacent, we must choose four of these five spaces in which to place a male student. We must also choose which four male students occupy the front row, which can be done in $$\binom{8}{4}$$ ways. These four male students can be arranged in the front row in $$4!$$ ways.

Since we have chosen which students occupy the front row, the remaining students must occupy the back row. There are $$4!$$ ways to arrange the girls in the back row and $$\binom{5}{4}4!$$ ways to arrange the boys in the back row.

Hence, there are $$\binom{8}{4}4!\binom{5}{4}\binom{8}{4}4!4!\binom{5}{4}4! = 5^2 \cdot 8!8!$$ seating arrangements of four male and four female students in which no two male students in the same row sit in adjacent seats.

Under the interpretation that no two male students in the same row or same column can be adjacent:

If no two male students can be adjacent to each other in the same row or in the same column, then the number of admissible seating arrangements is instead $$\binom{8}{4}4!\binom{5}{4}\binom{8}{4}4!4!4! = 5 \cdot 8!8!$$ since each male student in the second row would have to sit behind a female student in the first row and each female student in the second row would have to sit behind a male student in the first row, in which case there are $$4!$$ ways to seat the male students in the four columns occupied by female students in the first row and $$4!$$ ways to seat the female students in the four columns occupied by male students in the first row.

For males to sit apart, # in a row can't exceed $$4\,$$, hence exactly $$4$$ male and $$4$$ females in each row:$$\quad-F-F-F-F-$$

The $$4$$ males can be placed in the $$5\;$$ blanks in each row in $$C_4^5 =5$$ ways, $$\;$$in the two rows in $$\;5\cdot5 = 25$$ ways

Permute the males and females in the seats filled by them to give
total arrangements as $$\;25\cdot8!8!$$