Permutation Question

Question:
Six natives and two foreigners are seated in a compartment of a railway carriage with four seats either side. In how many ways can the passengers seat themselves if
a) the foreigners do not sit opposite each other,
b) the foreigners do not sit next to each other?


a) 6 natives 2 foreigners

the space is like
_ _ _ _
_ _ _ _
the foreigners sit opposite each other
f1 _ _ _ f2 _ _ _
f2 _ _ _ or f1 _ _ _
there are 2 ways of arranging the foreigners ; 2!
the foreigners can sit opposite each other in four ways

the six natives can be arrange in 6! ways to fill the six spaces left
thus the answer is 2! x 4 x 6! = 34560

b)
I am having problems with this one
my idea is to find out the number of ways in which the two foreigners are seated together and then subtract this from the total number of ways of them seating with no restriction.

Suppose the two foreigners are considered as a single entity then there are 7 objects. therefore ti can be arranged in 7! ways and the two foreigners can be arranged in 2! ways.
thus; 2! x 7! ways for the two foreigners can sit together

now to find the real answer we do
8! - (2! x 7!) = 30240
but the book gives an answer of 31680

Please help me what did i do wrong for part b

You've put the two benches together so over counted slightly.
The two benches of size 4 are on opposite sides of the carriage so there are not 7 ways of sitting together.

sorry but i dont understand
can you explain it a bit more


thank you

You seem to assume you have a single bench of 8, so there are 7 ways of sitting next to each other.
There are two benches of four seats each. There are not 7 ways of putting the two people next to each other.
You are c;lose to the right answer.

For A) why isn’t it 8!-(4 x 3 x 2 * 6!)?

Surely there are 4 x 3 = 12 combinations for the two foreigners?

Well done on getting the answer, perhaps the OP could say where the question comes from?
For A) there are only two foreigners. Each seat on one side could have one of the foreigners sat in it. The opposite seat would be uniquely determined by the other foreigner. So there are
4*2*6!
Ways to subtract. I'm not sure why you included the extra 3?

Note these questioms are less about applying a standard P/C formula, rather showimg you understand where they come from and getting your hands dirty, though ypu have to be very careful with the logic.

Also there was a couple of books on the ukmt site related to this. One simpler, then a follow on. Have a look and let me know if you can'tfind them.

Okay now I break down the logic I realise my mistake I believe.
Instead of counting the illegal combinations for the foreigners, I was calculating the legal ones so I was mixing it all up. I get my error now.
I see http://shop.ukmt.org.uk/art-of-problem-solving/introduction-to-counting-and-probability-2-book-set these two, but it’s £35 and I don’t think it completely relates to my a level content. Do you think I should work through it and buy it just to develop my general maths skills? Or should I purchase a book that’s more related to my a level course and have the benefits of learning a level maths and developing maths skills too.

I forgot you were focussing on a level content. They wouldn't really be appropriate.

Yeah okay. Final question.
For part A) , if you find the legal combinations without using the illegal questions I get that, you can have any 4 seats on the top row for the foreigner, then for each of those 3 other seats on the second row. Then the foreigners can be either way round so x 2. And then the 6 locals in any permutation of the 6 remaining seats

So I get this, 4 x 3 x 2! x 6! = 17280
It seems the actual answer is 4 x 3 x 2! x 6! x 2= 34560.
Why is it an extra x 2? Is it actually 3! Not 3? If so why. I can’t see what I’m missing.

Only considering legal combinations is actually very similar.
You have 8 possibilities for the first foreigner.
Then there are 6 possibilities for the second foreigner.
Then 6! For the locals so 34560.

You missed the fact that there are two rows for the first foreigner.

Question:
Six natives and two foreigners are seated in a compartment of a railway carriage with four seats either side. In how many ways can the passengers seat themselves if
a) the foreigners do not sit opposite each other,
b) the foreigners do not sit next to each other?

Question taken from The Brexit Party's Big Book Of Sums For English Youngsters

Question taken from The Brexit Party's Big Book Of Sums For English Youngsters

Nah, there arn't any foreigners in that :-).

You seem to assume you have a single bench of 8, so there are 7 ways of sitting next to each other.
There are two benches of four seats each. There are not 7 ways of putting the two people next to each other.
You are c;lose to the right answer.

What level question would this be? Year 1 or year 2 a level maths? Sorry for the obnoxious question

Yeah okay. Final question.
For part A) , if you find the legal combinations without using the illegal questions I get that, you can have any 4 seats on the top row for the foreigner, then for each of those 3 other seats on the second row. Then the foreigners can be either way round so x 2. And then the 6 locals in any permutation of the 6 remaining seats

So I get this, 4 x 3 x 2! x 6! = 17280
It seems the actual answer is 4 x 3 x 2! x 6! x 2= 34560.
Why is it an extra x 2? Is it actually 3! not 3? If so why. I can’t see what I’m missing.

Only considering legal combinations is actually very similar.
You have 8 possibilities for the first foreigner.
Then there are 6 possibilities for the second foreigner.
Then 6! For the locals so 34560.

You missed the fact that there are two rows for the first foreigner.

sorry but what do you mean by illegal and legal combinations