Permutation Question

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#1
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

0
1 year ago
#2
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.

(Original post by bigmansouf)
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

0
#3
sorry but i dont understand
can you explain it a bit more

thank you
(Original post by mqb2766)
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.
0
1 year ago
#4
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.
(Original post by bigmansouf)
sorry but i dont understand
can you explain it a bit more

thank you
1
1 year ago
#5
Omg I actually got B) I’m so proud of myself! Jesus Christ. Do you get it now? I won’t say how you get it though.
0
1 year ago
#6
(Original post by mqb2766)
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
0
1 year ago
#7
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?
Last edited by Maximus 190; 1 year ago
0
1 year ago
#8
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.

(Original post by Maximus 190)
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?
Last edited by mqb2766; 1 year ago
0
1 year ago
#9
(Original post by mqb2766)
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-probl...ity-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.
0
1 year ago
#10
I forgot you were focussing on a level content. They wouldn't really be appropriate.
(Original post by Maximus 190)
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-probl...ity-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.
0
1 year ago
#11
(Original post by mqb2766)
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.
Last edited by Maximus 190; 1 year ago
0
1 year ago
#12
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.

(Original post by Maximus 190)
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.
0
1 year ago
#13
(Original post by mqb2766)
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.
Oh yes of course they don’t have to sit in opposite rows only! Silly.
It still seems to work ignoring that there are two rows for the second foreigner as I multiplied by 2! to have it in either orientation, but your way is more intuitive and straight forward. (4 x 6 x 2! x 6!) vs (8 x 6 x 6!)

0
1 year ago
#14
(Original post by bigmansouf)
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
0
1 year ago
#15
Nah, there arn't any foreigners in that :-).

(Original post by the bear)
Question taken from The Brexit Party's Big Book Of Sums For English Youngsters
2
1 year ago
#16
(Original post by the bear)
Question taken from The Brexit Party's Big Book Of Sums For English Youngsters
Lmao, the discrete propaganda.
1
1 year ago
#17
(Original post by mqb2766)
Nah, there arn't any foreigners in that :-).
They’re in the other train going down to the prison
Last edited by Maximus 190; 1 year ago
0
#18
(Original post by mqb2766)
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.
(Original post by Maximus 190)
What level question would this be? Year 1 or year 2 a level maths? Sorry for the obnoxious question
This question is from a maths book by J.B. Backhouse, S.P.T. Houldsworth Pure Mathematics 1
This question is under permutation & combinations
it is mostly likely under statistics year 1 i think

i found the answer it is 31680
I found it this way

2 foreigners (F for foreigners) 6 natives (N for natives)

if the two F sit together - they can be arranged in 2! ways

there are two benches of 4 seats as pointed by mqb2766

so for the first bench;
they can sit in the first two seats or the 2nd and 3rd seats or the 3rd seats and 4th seats - this makes it 3 ways
for the second bench;
they can sit in the first two seats or the 2nd and 3rd seats or the 3rd seats and 4th seats - this makes it 3 ways
in total that makes it 6 ways

you can arrange the natives in 6! ways

thus the two foreigners can sit together in (2! x 6 x 6!) ways

with no restriction it is 8! ways

the answer will be; 8! - (2! x 6 x 6!) = 31680
1
#19
sorry but what do you mean by illegal and legal combinations
(Original post by Maximus 190)
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.
(Original post by mqb2766)
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.
0
1 year ago
#20
Just allowed / not allowed.

(Original post by bigmansouf)
sorry but what do you mean by illegal and legal combinations
0
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