There are 10 seats numbered 1 through 10. There are 5 boys and 4 girls. The girls must sit in the even numbered seats.

Question

There are 10 seats numbered 1 through 10. There are 5 boys and 4 girls. The girls must sit in the even numbered seats. How many ways can they sit such that there are no empty seats?

Here is what I tried:

  g b g b g b g b b
1 2 3 4 5 6 7 8 9 10

or

  b b g b g b g b g
1 2 3 4 5 6 7 8 9 10

Therefore the girls can be arranged 4! but in two separate configurations so: 4!*2 accounts for all the ways the girls can sit.

I then used the same logic for boys (it does not state that boys can not sit in the even seats, so a boy can sit in the seat numbered "2"): 5!*2

I multiplied these results together: 48*240=11,520

There is no answer guide for this so I am just trying to see if my thought process is correct.

Thanks!

Answer 1

The number of ways to seat the girls is $4! \binom{5}{4}$ (5 available seats, select 4, permute 4); the number of ways of seating the 5 boys is $5! \binom{6}{5}$ (after seating the girls, there are 6 seats free; select 5 for the boys, permute). In all:

$\begin{align*} 4! \binom{5}{4} \cdot 5! \binom{6}{5} &= 4! \frac{5!}{4! 1!} \cdot 5! \frac{6!}{5! 1!} \\ &= 5! \cdot 6! \\ &= 86\,400 \end{align*}$

Answer 2

Supposing the seat 1 will not be empty, the number of ways to seat the girls is $4!\cdot\binom{5}{4}$ and the number of ways to seat the boys is $5!$, so in total it is $4!\cdot\binom{5}{4}\cdot5!$.

Supposing now the seat 1 will be empty, you have the same number of ways to seat the boys and since there are only four even seats there are only 4! ways to seat the girls, therefore in total you have $4!\cdot\binom{5}{4}\cdot5!+4!\cdot5!$

Answer 3

The question was poorly worded.

There are ten seats numbered 1 through 10. There are five boys and four girls. The girls must sit in the even-numbered seats. How many ways can the nine children be seated?

This question has been correctly answered by vonbrand.

However, it seems the intended question was:

There are ten seats numbered 1 through 10. There are five boys and four girls. The girls must sit in the even-numbered seats. How many ways can the nine children be seated if there are no empty seats between people?

If there are no empty seats between people, then the nine children must sit in consecutive seats. Consequently, either seat 1 is empty or seat 10 is empty.

Case 1. Seat 1 is empty.

Since the girls must be seated in even-numbered seats, there are five seats available to them. Choose which four of these five seats will be occupied by the girls and arrange them in these seats. The remaining five seats must be occupied by the five boys. Arrange the boys in these seats. This can be done in $$\binom{5}{4}4!\binom{5}{5}5! = 5 \cdot 4! \cdot 5! = 5!5!$$ ways.

Case 2. Seat 10 is empty.

Since the girls must be seated in even-numbered seats, they must occupy all four of the remaining four even-numbered seats. Arrange them in these seats. The five boys must occupy the five odd-numbered seats. Arrange them in these seats. This can be done in $$\binom{4}{4}4!\binom{5}{5}5! = 4!5!$$ ways.

Total. Since the two cases are mutually exclusive and exhaustive, the number of admissible seating arrangements is $$5!5! + 4!5!$$