My Math Forum Number of arrangement(Probability)

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 May 28th, 2011, 05:25 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Number of arrangement(Probability) 1.Ten people go to cinema and sit in a group of ten reserved seats in the front row.If three of the ten people are sisters and must be seated together a)find the number of possible seating arrangement for all ten people b)find the number of possible seating arrangement for all ten people,if two of the remaining seven refuse to sit together My solution a) consider the 3 sister are 1 object,so total is 8!, but they can exchange their place so 8! x 3! b) this one i did wrong.and the solution teacher given is 8!(3!)- 7!(3!)(2!) without any explanation.So i don understand about it.I need help and explanation for this part only. Thank you.
 May 28th, 2011, 09:57 PM #2 Member   Joined: May 2011 Posts: 51 Thanks: 0 Re: Number of arrangement(Probability) Two guys don't seat together, then how many ways they will seat together... Consider 3 girls as 1(as u already did), also consider those 2 guys as 1 Now there are 7 people total so total arrangements are 7!*3!*2!(the same way as 'a') The rest arrangements are where those two will not seat together = 8!*3!-7!*3!*2!
 May 29th, 2011, 06:38 AM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Number of arrangement(Probability) Ok,so do you mean that in the 7 people consists of the 3 girls consider as 1 and 2 guys consider as 1 ,so the arrangements is 7!3!2!.
June 4th, 2011, 04:38 PM   #4
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Quote:
 Originally Posted by hoyy1kolko 1.Ten people go to cinema and sit in a group of ten reserved seats in the front row.If three of the ten people are sisters and must be seated together a)find the number of possible seating arrangement for all ten people b)find the number of possible seating arrangement for all ten people,if two of the remaining seven refuse to sit together
a. I got 8*7!.

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### what is the probability that the 3 sisters all sit together

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