My Math Forum Permutations Help

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 October 4th, 2016, 05:51 PM #1 Newbie   Joined: Oct 2016 From: Canada Posts: 1 Thanks: 0 Permutations Help Hello, I'm stuck on the following problem and am hoping for some guidance: For part (a), I believe the answer should be 10!. For (b, i), I'm thinking I should do 10*9*8*7*6*5 for distributing the 10 seats about the 6 people there, giving 151,200. It's at (b, ii) that I get stumped. At first I was thinking of putting the 3 friends together (which can be done in 3! ways). Then, because I put the 3 friends together, I would have 3 other friends left, so 3+1 = 4 groups. Among these 4 groups, the 8 seats can be distributed by 8*7*6*5. When I multiple 3!*8*7*6*5 together, I get 10,080, which is nowhere close to the correct answer of 33,600. I would really appreciate it if someone could tell me where I'm going wrong. Thank you!
 August 7th, 2017, 09:36 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,243 Thanks: 1439 Do you still need help on this?
August 8th, 2017, 01:50 AM   #3
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 Originally Posted by purplesheep Hello, I'm stuck on the following problem and am hoping for some guidance: For part (a), I believe the answer should be 10!. For (b, i), I'm thinking I should do 10*9*8*7*6*5 for distributing the 10 seats about the 6 people there, giving 151,200. It's at (b, ii) that I get stumped. At first I was thinking of putting the 3 friends together (which can be done in 3! ways). Then, because I put the 3 friends together, I would have 3 other friends left, so 3+1 = 4 groups. Among these 4 groups, the 8 seats can be distributed by 8*7*6*5. When I multiple 3!*8*7*6*5 together, I get 10,080, which is nowhere close to the correct answer of 33,600. I would really appreciate it if someone could tell me where I'm going wrong. Thank you!
For b)ii), Alice has only 8 seats to choose from (if she chooses an outermost seat, the requirement of being seated next to two friends won't be satisfied). Thus there are 5*4 ways Alice can sit next to two friends.

The remaining 3 friends can sit in any of the other seats. i.e., the first friend can choose to sit in any 7 seats, the second can choose from 6 seats and the third from 5 seats. There are (7-0)(7-1)(7-2) = 210 ways they can be seated.

Altogether, there are 8*20*210 = 33600 ways.

Last edited by skipjack; August 8th, 2017 at 04:51 AM.

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