Germs and bombs (1 Viewer)

[HeroicPandas]

Eight people attend a meeting. They are provided with 2 circular tables, one seating 3 humans, the other 5 humans.

i) How many seating arrangements are possible?
ii) If the seating is done randomly, what is the probability that a particular couple are on different tables?

I will be absolutely grateful if any human in boredofstudies, assists me in this question from 2000 SBHS (Q6)

PS. im bad at perms and commbs

 

[nightweaver066]

i) Selecting 3 humans first, so 8C3. Arranging that table by multiplying by 2! so we now have 8C3 * 2!

We are left with 5 humans for the other table. Arranging that table by multiplying by 4! so we now have 8C3*2!*4! which should be our answer.

ii) P(particular couple on different table) = 1 - P(particular couple on same table)

We consider the cases where the particular couple is on the table seating 3 humans, and then the other table that seats 5 humans.

1st case - 3 human table - Place the couple there. Now select 1 human from the remaining 6, 6C1, arrange this table, 2! and now you have 6C1*2!
Now arrange the other table, 4!, so you have 6C1*2!*4!

2nd case - 5 human table - Place the couple there. Now select 3 humans from the remaining 6, 6C3, arrange the table, 4! and now you have 6C3*4!
Now arrange the other table, 2!, so you have 6C3*4!*2!

Total possible seating arrangements was 8C3*2!*4!, so P(particular couple on different tables) = 1 - (6C1*2!*4! + 6C3*4!*2!)/(8C3*2!*4!) = 15/28

this better be right

 

[braintic]

i) Selecting 3 humans first, so 8C3. Arranging that table by multiplying by 2! so we now have 8C3 * 2!

We are left with 5 humans for the other table. Arranging that table by multiplying by 4! so we now have 8C3*2!*4! which should be our answer.

Even though I'm sure that this was the intended answer, I've always utterly disagreed with this solution to this type of question.
The fact that there are two tables completely removes the symmetry that allows us to use (n-1)! instead of n!.
That is, keeping one table fixed, and rotating the people at the other table by one space creates a completely new arrangement from the point of view of both the people sitting at the tables, and from an outside observer.

 

[HeroicPandas]

i) Selecting 3 humans first, so 8C3. Arranging that table by multiplying by 2! so we now have 8C3 * 2!

We are left with 5 humans for the other table. Arranging that table by multiplying by 4! so we now have 8C3*2!*4! which should be our answer.

ii) P(particular couple on different table) = 1 - P(particular couple on same table)

We consider the cases where the particular couple is on the table seating 3 humans, and then the other table that seats 5 humans.

1st case - 3 human table - Place the couple there. Now select 1 human from the remaining 6, 6C1, arrange this table, 2! and now you have 6C1*2!
Now arrange the other table, 4!, so you have 6C1*2!*4!

2nd case - 5 human table - Place the couple there. Now select 3 humans from the remaining 6, 6C3, arrange the table, 4! and now you have 6C3*4!
Now arrange the other table, 2!, so you have 6C3*4!*2!

Total possible seating arrangements was 8C3*2!*4!, so P(particular couple on different tables) = 1 - (6C1*2!*4! + 6C3*4!*2!)/(8C3*2!*4!) = 15/28

this better be right

Thank you so much for ur colossal assistance! (i get everything)

The horrible solutions displayed different answers

 

[braintic]

Thank you so much for ur colossal assistance! (i get everything)

The horrible solutions displayed different answers

The problem with the first answer (as is the problem with many answers in this topic) is that there is no definition as to what a unique arrangement is. The supplied answer to no 1 is correct if all you care about is which table a particular person sits at, and not their actual position at that table. It is clear that this is what they wanted only after reading part (ii).

However there is no way to get their answer to part (ii) from their wording.

 

[Girls]

Even though I'm sure that this was the intended answer, I've always utterly disagreed with this solution to this type of question.
The fact that there are two tables completely removes the symmetry that allows us to use (n-1)! instead of n!.
That is, keeping one table fixed, and rotating the people at the other table by one space creates a completely new arrangement from the point of view of both the people sitting at the tables, and from an outside observer.

Very interesting point. Will discuss this with my maths teacher.