anonymous
  • anonymous
In a doctor's waiting room, there are 14 seats in a row. Eight people are waiting to be seated. Three of the 8 are in one family, and want to sit together. How many ways can this happen?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I don't know how to do this, could someone explain it to me?
anonymous
  • anonymous
@ganeshie8 ?
anonymous
  • anonymous
They could sit in the first 3 seats so the last person is sitting in seat three or they could move down one and the last person could be sitting in seat 4. Since the first two people are always before the third take out two seats so you now have 12 seats and one person. how many ways can one person sit in 12 seats? so the answer is 12

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More answers

anonymous
  • anonymous
@abuehler you seem to have missed the point of the problem.
rsadhvika
  • rsadhvika
3! x 12 P 6
rsadhvika
  • rsadhvika
imagine we tie the 3 people from same family as 1. we have 5 more people who can sit as they wish Now, overall we have 6 items to be placed. and out of these 1 item will occupy 3 seats. so 6 items to be placed and 12 places to be filled
rsadhvika
  • rsadhvika
that gives you 12P6 and since the 3 ppl we tied can permute among themselves, you must multiiply above wid 3!
anonymous
  • anonymous
thank you, that makes sense now.
rsadhvika
  • rsadhvika
good, here we assumed that ur professor is not a psycho... and he simply wants the family to stick together without empty chairs between them - the simple case ;)
anonymous
  • anonymous
I knew there was a simple solution I had missed
rsadhvika
  • rsadhvika
it gets bit complex, if we allow empty chairs to exist between the family, you see wat i mean ? :)
anonymous
  • anonymous
yes, I see. That would be messy.
anonymous
  • anonymous
So is this resolved or not?
anonymous
  • anonymous
Here is my strategy: 1) Assign the family their seats first. 2) Assign the remaining people there seats.
anonymous
  • anonymous
1a) The three family members are together, members aren't distinct: Sit them all to the far left edge. They can scoot over \(14-3=11\) times. So including their current seating, there are \(12\) ways to seat them where the family members are indistinct. 2a) Making 1a) distinct This is just a matter of permutations. Three people are permuted among three seats: \(^3P_3=3!\) This means that for 1) we get \(12\times 3!\)
anonymous
  • anonymous
2) Now that we know \(3\) seats are taken, there are \(14-3=11\) seats left and \(8-3=5\) people left. So we permute \(11\) seats among \(5\) people: \(^{11}P_5=11!/6!\).
anonymous
  • anonymous
Combining these two, I get:\[ 12\times 3!\times \frac{11!}{6!} \]
anonymous
  • anonymous
What?
rsadhvika
  • rsadhvika
5 members to fill 11 seats left 11!/(11-5)!
anonymous
  • anonymous
Are you guys challenging my answer or...?
rsadhvika
  • rsadhvika
our both answers are exact same, no question of challenging :)

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