One question (1 Viewer)

[kooltrainer]

a nine member fund raising commitee consists of 4 students, 3 teachers and 2 parents. The committee meets around a circular table.


a)how many different arrangements of the 9 members around the table are possible if the students sit together as a group and so do the teachers, but no teacher sits next to a student?


b)One student and one parent are related. Given that all the arrangements in (a) are equally likely, what is the probability that these 2 members sit next to each other?

 

[Pwnage101]

i'm not too sure about this, but for a) ill have a go

treat them as 2 separate groups and 2 individuals, since there ar eonly 2 parents, these have to separate the 2 groups, so there is only 2 WAYS TO UNIQUELY ARRANGE THE FOUR 'PARTIES' IN A CIRCLE (PARTIES = parent1, parent2, students, tachers)

but there are 4 factorial ways of arranging students WITHIN that group, and tehre are 3 factorial ways of arranging teachers WITHIN that group

Not completely sure if i'm right (im probs wrong) but i think
a). = 2X4!X3! = 288????

i'll have a very rough gues of b). i think it's either 1/12, 1/8, or 1/4, but thats very rough!????

 

[lolokay]

a) 4!3!2! = 288
b)2/4*1/2 = 1/4

is this really mx2? it seems easy even for mx1 (assuming I'm correct)

 

[bored of sc]

I understand this isn't my question; I fully get a) but can someone explain their reasoning behind part b). Cheers.

 

[lyounamu]

I understand this isn't my question; I fully get a) but can someone explain their reasoning behind part b). Cheers.

If you draw the circle. There are 4 students with 2 parents surrounding them. So the chance are basically 2/4 (chance of being situated outside the group) x 1/2 (chance of having a right parent out of two) = 1/4

I also doubt that this question could be 4 Unit. I didn't even know that Permutations was part of the 4 Unit topic (or is this harder 3 Unit topic?)

 

[lyounamu]

i think you guys are missing out the rotation-factor for part a)

4! x 3! x 2! x 9 = 2592 (9 comes from the rotations)

Just realised that it's redundant from the later posts.

 

[kooltrainer]

i think you guys are missing out the rotation-factor for part a)



View attachment 16711

4! x 3! x 2! x 9 = 2592 (9 comes from the rotations)

i thought u dun count the rotations.. because they sit at the same place no matter which way u rotate.. so its redundant?

 

[solomarc20]

I also doubt that this question could be 4 Unit. I didn't even know that Permutations was part of the 4 Unit topic (or is this harder 3 Unit topic?)

Permutations questions have turned up in the 4U paper on a number of occasions. However, this question is not likely 2b a 4U question

 

[lolokay]

3! x 3! x 2! x 9 / 2592 = 1/4

or, 2/4 chance of being next to a parent, 1/2 chance of this being their parent