1. ## Seating arrangements.

quite confused with this particular question...

How many ways can 9 students(5 girls and 4 boys) be seated in a row if

there are no restrictions and if the first 3 must be boys?

i think its the difference in number of boys and girls that is confusing me

2. When there is no restriction, the possible number of ways is simply 9!.
When the first 3 must be boys, these 3 boys can sit in 3! ways and the rest 6 can sit in 6! ways. Hence the total number of possible ways in this case is 3!*6!.

3. ok i used what you showed and got farther into the question, but am not stuck at alternating seating girl then boy AND boy then girl?
would it be 5!*4!*9!? does it change if going from girl > boy instead of boy >girl?

4. Hello, slowlearner!

How many ways can 5 girls and 4 boys be seated in a row
if the first 3 must be boys?

Select three of the four boys: .$\displaystyle {4\choose3} \,=\,4$ choices.
Seat them in the first three chairs: .$\displaystyle 3! \,=\,6$ ways.
Seat the other six children: .$\displaystyle 6! = 720$ ways.

Therefore, there are: .$\displaystyle 4\cdot 6\cdot 720 \:=\:17,\!280$ ways.

5. How would i find out how many ways the 5 girls and 4 boys could be seated if the seating must alternate girl then boy? im quite confused here, is it 5!4!(9*8*7*6*5)? that is just my shot in the dark, im not quite sure how to solve this scenario

6. Originally Posted by slowlearner
How would i find out how many ways the 5 girls and 4 boys could be seated if the seating must alternate girl then boy? im quite confused here, is it 5!4!(9*8*7*6*5)? that is just my shot in the dark, im not quite sure how to solve this scenario
Note that the way you have written the question the row must begin with a girl.
So the answer is $\displaystyle (5!)(4!)$.
Sit a girl in ever other seat, then put a boy between them.

7. thank you! so if you were to seat it boy then girl then it would just be (4!)(5!)? or can this not be done since it would end up being girl next to girl at the end due to only 4 boys?

8. Originally Posted by slowlearner
thank you! so if you were to seat it boy then girl then it would just be (4!)(5!)? or can this not be done since it would end up being girl next to girl at the end due to only 4 boys?
But then you have two girls sitting together at the last two seats. Do you allow that?

9. thats what i was asking myself :P its got me quite confused. probability and statistics have been killing me all semester!

10. Originally Posted by slowlearner
thats what i was asking myself :P its got me quite confused. probability and statistics have been killing me all semester!
It is not "probability and statistics have been killing" you but the meaning of the language. It says that the girls and boys must alternate.
So it got be $\displaystyle gbgbgbgbg$.

11. i agree! i think it could possibly be 6! * 4! * 3 * 2 * 1?

12. Originally Posted by slowlearner
i agree! i think it could possibly be 6! * 4! * 3 * 2 * 1?
That is wrong.

13. Originally Posted by Plato
That is wrong.
can you elaborate?

14. Originally Posted by slowlearner
can you elaborate?