Question

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?

Answer 1

I don't know how to do this, could someone explain it to me?

Answer 2

@ganeshie8 ?

Answer 3

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

Answer 4

@abuehler you seem to have missed the point of the problem.

Answer 5

3! x 12 P 6

Answer 6

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

Answer 7

that gives you 12P6
and since the 3 ppl we tied can permute among themselves, you must multiiply above wid 3!

Answer 8

thank you, that makes sense now.

Answer 9

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 ;)

Answer 10

I knew there was a simple solution I had missed

Answer 11

it gets bit complex, if we allow empty chairs to exist between the family, you see wat i mean ? :)

Answer 12

yes, I see. That would be messy.

Answer 13

So is this resolved or not?

Answer 14

Here is my strategy:
1) Assign the family their seats first.
2) Assign the remaining people there seats.

Answer 15

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 \(12\times 3!\)

Answer 16

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: \(^{11}P_5=11!/6!\).

Answer 17

Combining these two, I get:\[
12\times 3!\times \frac{11!}{6!}
\]

Answer 18

What?

Answer 19

5 members to fill 11 seats left 11!/(11-5)!

Answer 20

Are you guys challenging my answer or...?

Answer 21

our both answers are exact same, no question of challenging :)