perms/combs q (1 Viewer)

[scardizzle]

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

i) How many arrangements are possible?

ii) If the seating is done randomly, what is the probability that a particular couple are on a different table?

for (i) i did this:

choose 3 people from 8 x no. of arrangements for each table

= 8C3 x 2! x 4!

but the answers say the solution is 8!/(3!5!)

???

also for ii) P(not sitting together) = 1 - P(sitting together)

1- [ 2C2 x 6C1 x 2! x 4! + 2C2 x 6C3 x 2! x 4!]/answer (i)

and this is wrong as well

so what am i doing wrong?

 

[khorne]

For the first one, they are just arranging 8 people around both tables (at the same time) and then canceling repetitions at both tables by dividing by 3 and 5

 

[scardizzle]

but why are there 3! x 5! repetitions?

and also why is my method incorrect?

 

[khorne]

I edited original post...Shouldn't be factorial...just 3 and 5, which makes your method correct also...Nothing wrong with it.

 

[scardizzle]

okay thanks alot for clearing that up for me