Combinatorics questions.

[Sonal7]

I have one last Q on combinatorics.



I have one last Q (I hope) on the topic and I am stuck on part (c). The answer seems obvious but I disagree slightly with the answer book. I think there are 4 possible positions that Alan and Tina may take, its top left or right, or bottom left or right. Since they can arrange themselves. So the total arrangements are 8 x 9! if you think of A and T as a container. But the book says answer is 8 x 8!. Its a slight disagreement. I am pretty sure my answer is wrong ! I wish I had not asked such a pedantic question. However I am completely unsure about part (d). I think the probability of part (c) divided by total arrangements. I think its 8*8! divided by 12!/2!, my answer is 2/1485

 

[Dr.Peterson]

Please explain your choice of 9! rather than 8!. What are you thinking? (There are 8 people to arrange in 8 seats.)

 

[Sonal7]

Please explain your choice of 9! rather than 8!. What are you thinking? (There are 8 people to arrange in 8 seats.)

Yes but I am thinking that the two who sit together count as another person but i am wrong. I am ok now that we know that there are 8 position for the two to sit together and the rest need to be arranged.

 

[Dr.Peterson]

They sit together, but not in the same seat; and there are still only 8 other people.

When you count a group of people as one in part of a problem, you have to keep the reality of the problem in mind so you don't forget aspects in which they are treated separately.

 

[pka]

Yes but I am thinking that the two who sit together count as another person but i am wrong. I am ok now that we know that there are 8 position for the two to sit together and the rest need to be arranged.

There twelve pairs of adjacent seats. Choose a pair, seat the two either BG or GB: \((\mathcal{C}^{12}_1=12)\cdot(2)\)
That leaves ten unoccupied seats for eight other people: \((\mathcal{P}^{10}_8=\dfrac{10!}{2!})\).
\((\mathcal{C}^{12}_1)\cdot(2)\cdot(\mathcal{P}^{10}_8)\)

 

[Dr.Peterson]

There twelve pairs of adjacent seats. Choose a pair, seat the two either BG or GB: \((\mathcal{C}^{12}_1=12)\cdot(2)\)
That leaves ten unoccupied seats for eight other people: \((\mathcal{P}^{10}_8=\dfrac{10!}{2!})\).
\((\mathcal{C}^{12}_1)\cdot(2)\cdot(\mathcal{P}^{10}_8)\)

I don't think you're looking at part (c), which is under discussion.