# Thread: Difficult (for me, that is) Combinatorics/Probability problem

1. Originally Posted by aman_cc
Can u plz explain Q2 and Q3 - I have not really followed them. Also it would be good if u cld share your working
Okay, sure.

Question 2 - this means that if one person turns up at 9am (ie he has the possibility of getting any of the interview slots - he doesn't necessarily have to have his interview at 9am)... what is the chance that he will have his second interview right away after his first one. ie: he will have his interviews at 9-9:15/9:15-9:30 OR 9:15-9:30/9:30-9:45 OR 9:30-9:45/9:45-100. Remembering that he can have his 9am interview with either person etc.

Question 3 - I think this is basically the same method. But this person turns up at 9am (doesn't have to have their interview then) but has to be finished before 9:50 - ie this means what is the probability they will have both their interviews with in the first 3 time slots (they can't have an interview from 9:45-10:00) since they need to be somewhere else at this time.

I got 1/2 for both of them.

My working for question two was....

well.... if you fix the person in a position (say position one) in interview room one, there is 3! ways of organizing everyone else's interview in this room. Now, there are 9 derrangements for the second room (we got this from question one) but there are only 3 that fix this person in the next time slot (position 2) (i got this because i wrote all the 9 derrangements out). So multiply 3! x 3. Also however, there are 3 possible ways of him getting consecutive interviews. 1-2, 2-3, 3-4. So multiply by 3 again... so at the moment we have 3! x 3 x 3. However, because this is only going one way - room 1, then room 2. we can obviously go the other way, room 2 to room 1. So there are two possibly ways of organising that. So x 2. Therefore all up we have 3! x 3 x 3 x2 = 108. Now there are 216 possible combinations... so we have 108/216 = 1/2. What do you think?

For question 3 i think its the same principle again. except she needs to finish her interviews before the 9:45 slot, so she needs to be in the first three interview rooms. The possible ways this can happen is: (1st interview at 9, second at 9:15), (1st interview at 9, second at 9:30), (1st interview at 9:15, second interview at 9:30) So dealing with these specific cases: Case one where her first interview is at 9. If you fix her in her first interview at 9... there's 3! ways of organising everyone else around this fixed person. Then, there are 6 possible derangements that fix this person in her second interview either at 9:15 or 9:30. Case two, her first interview at 9:15. There is 3! ways of organising everyone around this, then there are 3 derrangements that fix her to position 3 in interview room 2. So therefore we have (number of possibilities for case 1)+(number of possibilities for case 2) = (3!x6)+(3!x3)=36+18=54. But because you can go room 1 then room 2, or room 2 then room 1. We multiply by 2 which = 108. again 108/216 = 1/2.

So did you understand any of that?? haha, thanks again for all of your help!!

2. Q2 appears correct to me

And so does Q3 !!

Page 2 of 2 First 12