3u Trial Question :/ (1 Viewer)

[Ikki]

There was solutions for 1st part but not the 2nd.

Question:
10 people arrive at a restaurant. There are two circular tables. 6 people sit at one table and 4 in another.
i) How many different seating arrangements are possible using these tables?
ii) Assuming arrangement is random, what is the probability of a couple sitting together?

1st part solution was:
10C6 x 5! x 3! = 151200

EDIT:
Original

Original Question:

Ten people arrive to eat at a restaurant. The only seating available for them is at two circular tables, one that seats six persons, and another that seats four.
i) Using these tables, how many different seating arrangements are therefor the 10 people?
ii) Assuming that the seating arrangement is random, what is the probability that a particular couple will be seated at the same table.

 

[seanieg89]

It's quite sloppily written question (unless you are paraphrasing).

Are the tables circular?

Did each of the 10 people come with a partner?

Does "together" mean adjacent or "at the same table"?

Are we looking for the probability of at least one couple sitting together, or of all couples sitting together?

 

[Ikki]

Yep paraphrasing.
1. Yes
2. No.
3. I presume adjacent
4. One

Sorry for the confusion

 

[Kurosaki]

There was solutions for 1st part but not the 2nd.

Question:
10 people arrive at a restaurant. There are two circular tables. 6 people sit at one table and 4 in another.
i) How many different seating arrangements are possible using these tables?
ii) Assuming arrangement is random, what is the probability of a couple sitting together?

1st part solution was:
10C6 x 5! x 3! = 151200

A couple as in a single man and a single woman together right?
Are there 5 men and 5 women? Just post the original Q.

 

[Ikki]

K looked back and copied out the exact question. And no a couple is not a single man and a single woman, that is something that I must have included even in my paraphrasing.
Edited original post.
But yeah all good now, anyone?

 

[Kurosaki]

There was solutions for 1st part but not the 2nd.

Question:
10 people arrive at a restaurant. There are two circular tables. 6 people sit at one table and 4 in another.
i) How many different seating arrangements are possible using these tables?
ii) Assuming arrangement is random, what is the probability of a couple sitting together?

1st part solution was:
10C6 x 5! x 3! = 151200

EDIT:
Original

Original Question:

Ten people arrive to eat at a restaurant. The only seating available for them is at two circular tables, one that seats six persons, and another that seats four.
i) Using these tables, how many different seating arrangements are therefor the 10 people?
ii) Assuming that the seating arrangement is random, what is the probability that a particular couple will be seated at the same table.

I think it might be ?
Consider when one of the couple is at the 6 seater- there are 5 places left out of 9, so the chances of the other being at the same table is
For when one is at the 4 seater, 3 places left out of 9, so . Adding,
But this answer seems too much..

 

[Ikki]

I think it might be ?
Consider when one of the couple is at the 6 seater- there are 5 places left out of 9, so the chances of the other being at the same table is
For when one is at the 4 seater, 3 places left out of 9, so . Adding,
But this answer seems too much..

I was thinking more along the lines of:

to make use of i)

PS: Forgot how to space in latex.

 

[mathsfreak1010]

I was thinking more along the lines of:

to make use of i)

PS: Forgot how to space in latex.

im not really sure if this is right but perhaps using what Ikki said above

number of ways two couples can sit together:
(2c2*8C4*4C4*4!*2)+(2c2*8C2*6C6*2*2)= 3472

p(couple sitting together)= 3472/151200 = 31/1350

whats the answer anyway?

 

[Ikki]

I don't know maybe someone is good at these here which is why i posted it.

 

[iStudent]

I'm not that good at probability, but this is what I think..
I'm assuming the couples sit next to each other
Case 1: the couples sit on the table with 6 seats
=(8C4 x 4! x 2) x (3!) = 20160
Case 2: the couples sit on the table with 4 seats
= (8C2 x 2! x 2) x (5!) = 13440
P = (20160+13440)/151200 = 2/9
Tell me if I missed out on something ..

Edit:

number of ways two couples can sit together:
(2c2*8C4*4C4*4!*2)+(2c2*8C2*6C6*2*2)= 3472

p(couple sitting together)= 3472/151200 = 31/1350

My method is similar to yours, except I multiplied by 3! in case 1 and 5! in case 2 because you can still arrange the people on the other table.

Btw, I think the question doesn't assume it's adjacent (just means they're on the same table). So I think the number of ways should be:
Case 1: on table 6
8C4 x 5! x 3! = 50400
Case 2: on table 4
8C2 x 3! x 5! = 20160
P = 50400+20160/151200 = 7/15

 

[mathsfreak1010]

I'm not that good at probability, but this is what I think..
I'm assuming the couples sit next to each other
Case 1: the couples sit on the table with 6 seats
=(8C4 x 4! x 2) x (3!) = 20160
Case 2: the couples sit on the table with 4 seats
= (8C2 x 2! x 2) x (5!) = 13440
P = (20160+13440)/151200 = 2/9
Tell me if I missed out on something ..

Edit:

My method is similar to yours, except I multiplied by 3! in case 1 and 5! in case 2 because you can still arrange the people on the other table.

Btw, I think the question doesn't assume it's adjacent (just means they're on the same table). So I think the number of ways should be:
Case 1: on table 6
8C4 x 5! x 3! = 50400
Case 2: on table 4
8C2 x 3! x 5! = 20160
P = 50400+20160/151200 = 7/15

yes i did forget to multiply the other table arrangement so i think that way is correct.


oh yeah!- i over thought the questions lOL
it is just on the same table

 

[Ikki]

sounds about right Thanks guys.