The Student Room Group

Permutation question (i think) about boy and girl seating on a bench

Q: 8 boys and 2 girls sit on the bench. if the girls may sit neither at the ends nor together. In how many ways can they be arranged?

8 boys and 2 girls = 10 total
with no restriction 10!
neither at the ends nor together thus the events

a) not at the ends
for this part i believe that the two girls can sit next to each other
so there are 10 spaces on the bench
_ _ _ _ _ _ _ _ _ _
the two spaces at the ends are for the boys so
it is


B _ _ _ _ _ _ _ _ B
8 7
there are 6 boys & 2 girls left 8 in total; there are 6! ways of ordering the boys and 2! ways of ordering the girls

= 8 x 7 x 2 ! x 6! = 80640

b) not together

g g b b b b b b b b b

there are 2! ways of arranging the girls and 9! ways of arranging them all
so it will be 10! - (2! x 9!) = ways


since the events are mutually exclusive thus

80640 + 10! - (2! x 9!) = 2983680 ways


I have tried several times but i cannot get the answer 1693440

Please can you help
Is this question not asking for the event AnB?
Original post by bigmansouf
Q: 8 boys and 2 girls sit on the bench. if the girls may sit neither at the ends nor together. In how many ways can they be arranged?


Yes 10! is the unrestricted number... so now just take away the restriction numbers??

(A) How many arrangements if one girl is on the left end, and the other is on the right end?

(B) How many arrangements if both girls sit next to each other?

(C) How many arrangements if one girl is on the left end, and the other girl is elsewhere?

(D) How many arrangement is one girl is on the right end, and the other girl is elsewhere?




It would also be helpful if you stop posting new questions and respond to the people who try to help you in your threads in order for us to acknowledge whether we have helped you or not.
(edited 4 years ago)
Reply 3
Original post by RDKGames
Yes 10! is the unrestricted number... so now just take away the restriction numbers??

(A) How many arrangements if one girl is on the left end, and the other is on the right end?

(B) How many arrangements if both girls sit next to each other?

(C) How many arrangements if one girl is on the left end, and the other girl is elsewhere?

(D) How many arrangement is one girl is on the right end, and the other girl is elsewhere?




It would also be helpful if you stop posting new questions and respond to the people who try to help you in your threads in order for us to acknowledge whether we have helped you or not.

I ma truly sorry - I appreciate every help
my strategy was to go through all the questions I can and pile up all the questions i cannot do to one side and ask for help for each of them.
There are about 7 of them. I thought it will be better to ask all them in one go ( making threads on them) all in one go and then with the hints and help provided I will work through them through out the night and tomorrow.
I am truly sorry if tsr has rules against this method (if it can be seen as spamming) but I only thought of a more better way to suit me
Original post by bigmansouf
I ma truly sorry - I appreciate every help
my strategy was to go through all the questions I can and pile up all the questions i cannot do to one side and ask for help for each of them.
There are about 7 of them. I thought it will be better to ask all them in one go ( making threads on them) all in one go and then with the hints and help provided I will work through them through out the night and tomorrow.
I am truly sorry if tsr has rules against this method (if it can be seen as spamming) but I only thought of a more better way to suit me


Fair enough if that's how you want to do it, but I think tackling one question at a time is better... since you're not piled up and jumping between different contexts, and also all the helpers are there to help at the time they post, since they might not be around when you come back to the question thread later.
Reply 5
Original post by RDKGames
Fair enough if that's how you want to do it, but I think tackling one question at a time is better... since you're not piled up and jumping between different contexts, and also all the helpers are there to help at the time they post, since they might not be around when you come back to the question thread later.

thank you
Reply 6
Original post by RDKGames
Yes 10! is the unrestricted number... so now just take away the restriction numbers??

(A) How many arrangements if one girl is on the left end, and the other is on the right end?

(B) How many arrangements if both girls sit next to each other?

(C) How many arrangements if one girl is on the left end, and the other girl is elsewhere?

(D) How many arrangement is one girl is on the right end, and the other girl is elsewhere?




It would also be helpful if you stop posting new questions and respond to the people who try to help you in your threads in order for us to acknowledge whether we have helped you or not.

thank you very much i was able to answer the question
the difficult part for me was c and d

C) 8 boys & 2 girls

G on the left end can be done in 2! ways

the other girl cannot be allow to sit next to her or at the other end thus she is only allow to sit in 7 other places
then the rest of the boys can be arrange in 8! ways

= 2! x 7 x 8!

this is the same for part d

in the end
10! - ((2! x 7 x 8) + (2! x 7 x 8! ) + ( 2 x 8!) +(2 x 9!)) = 1693440
(edited 4 years ago)
Original post by bigmansouf
thank you very much i was able to answer the question

10! - ((2! x 7 x 8) + (2! x 7 x 8! ) + ( 2 x 8!) +(2 x 9!)) = 1693440


For a slightly different method:

We know all boys & girls must be allocated some position, so initially we could work out just the number of arrangements of boys and girls without regard to which specific boy or girl we are refering to.

Girls don't sit at the end so we have B(8 spaces)B

Then for the central 8 spaces we could allocate the boys/girls in 8C2 ways, but must subtract the number where the two girls sit together, 7C1.

Giving us 8C2-7C1 arrangements of boy/girl that meet the criteria.

Then, the specific boy/girl can be allocated into the arrangement, giving us 8! x 2! x (8C2-7C1) = 1693440
Original post by bigmansouf
I ma truly sorry - I appreciate every help
my strategy was to go through all the questions I can and pile up all the questions i cannot do to one side and ask for help for each of them.
There are about 7 of them. I thought it will be better to ask all them in one go ( making threads on them) all in one go and then with the hints and help provided I will work through them through out the night and tomorrow.
I am truly sorry if tsr has rules against this method (if it can be seen as spamming) but I only thought of a more better way to suit me

So, over the last couple of weeks, you've started almost 20 threads to do with permutation questions. It is starting to feel a bit excessive, TBH. It's also feeling like you really aren't progressing - but that maybe somewhat selection bias since you're only posting questions you're stuck on.

Some general thoughts:

A really common theme with these questions you're stuck on is that you haven't read the question carefully. To be fair, many are tricky questions, and it's easy to slip up. But if you know you're getting the wrong answer, you really ought to be able to go back and question your assumptions/interpretation of the question.

If you're stuck, you should see if there's a "smaller" version of the question you can manually verify. In this case, you could try what happens with only 2 boys and 2 girls. (And then possibly 3 boys and 2 girls). This would make it easier for you to see (and remember for the future) why your initial method has failed, and what you might need to change to fix things.

But to repeat myself: many of these questions are quite hard - so don't get too discouraged (and don't assume there's always going to be an "easy" way to answer a question - often it does seem there's nothing better than to break things down into quite a few cases).

I do slightly wonder why you're doing these questions: they seem rather hard for A-level.
Reply 9
Original post by DFranklin
So, over the last couple of weeks, you've started almost 20 threads to do with permutation questions. It is starting to feel a bit excessive, TBH. It's also feeling like you really aren't progressing - but that maybe somewhat selection bias since you're only posting questions you're stuck on.

Some general thoughts:

A really common theme with these questions you're stuck on is that you haven't read the question carefully. To be fair, many are tricky questions, and it's easy to slip up. But if you know you're getting the wrong answer, you really ought to be able to go back and question your assumptions/interpretation of the question.

If you're stuck, you should see if there's a "smaller" version of the question you can manually verify. In this case, you could try what happens with only 2 boys and 2 girls. (And then possibly 3 boys and 2 girls). This would make it easier for you to see (and remember for the future) why your initial method has failed, and what you might need to change to fix things.

But to repeat myself: many of these questions are quite hard - so don't get too discouraged (and don't assume there's always going to be an "easy" way to answer a question - often it does seem there's nothing better than to break things down into quite a few cases).

I do slightly wonder why you're doing these questions: they seem rather hard for A-level.

thank you just trying to improve in maths i want to forward to undertake a degree engineering or maths & computer science
If the 2 girls can't sit 2gether nor at the ends let us look at this question like this;
B1_B2_B3_B4_B5_B6_B7_B8
SO the girls can sit at any of those spaces In between the boys that way they can't sit 2gether nor can the sit at the end
So G1 has 7 ways she can sit and G2 has 6ways so 7×6=42
And the ways we can arrange the boys is 8!= 40320
So if you multiply the No. Of ways the girls n the boys can be arranged
40320×42=1693440

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