Probability Question Confusion: Seating 3 friends, 7 others in a row

plasmatic

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Question 7E-9 Three friends and seven other people are randomly seated in a row. Specify an appropriate sample space to answer the following two questions.

(a) What is the probability that the three friends will sit next to each other?

(b) What is the probability that exactly two of the three friends site next to each other?




Solution to 7E-9: Take as sample space the set of all possible orderings of the ten people on the ten seats. The sample space has 10! equally likely outcomes.

(a) The number of ways to choose three adjacent seats from the ten seats is 8. Hence the number of outcomes for which the three friends are seated next to each other is:


. . . . .\(\displaystyle 8\, \times\, 3!\, \times\, 7!\)

and so the probability that the three friends will sit next to each other is:

. . . . .\(\displaystyle \dfrac{8\, \times\, 3!\, \times\, 7!}{10!}\, =\, \dfrac{1}{15}\)

(b) The number of ways to choose two friends from the three friends is:

. . . . .\(\displaystyle \large{ \binom{3}{2} }\)

The number of outcomes for which the two chosen friends sit next to each other at one of the ends of the row and the third friend has a non-adjacent seat is:

. . . . .\(\displaystyle 2\, \times\, 7\, \times\, 2!\, \times\, 7!\)

while the number of outcomes for which the two chosen friends site next to each other but not at one of the ends of the row and the third friends has a non-adjacent seat is:

. . . . .\(\displaystyle 7\, \times\, 6\, \times\, 2!\, \times\, 7!\)

Hence the probability that exactly two of the three friends will sit next to each other is:

. . . . .\(\displaystyle \large{ \dfrac{\binom{3}{2}\, \left[\, 2\, \times\,7 \,\times\, 2! \,\times\, 7!\, +\, 7 \,\times\, 6 \,\times\, 2! \,\times\, 7!\, \right]}{10!}\, =\, \dfrac{7}{15} }\)



I don't understand how the answer to (b) was reached. I don't understand where any of the numbers come from in defining how many possibilities there are for each stated outcome. Could someone elaborate into how this was done?

Thanks.
 

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Question 7E-9 Three friends and seven other people are randomly seated in a row. Specify an appropriate sample space to answer the following two questions.

(a) What is the probability that the three friends will sit next to each other?

(b) What is the probability that exactly two of the three friends site next to each other?




Solution to 7E-9: Take as sample space the set of all possible orderings of the ten people on the ten seats. The sample space has 10! equally likely outcomes.

(a) The number of ways to choose three adjacent seats from the ten seats is 8. Hence the number of outcomes for which the three friends are seated next to each other is:


. . . . .\(\displaystyle 8\, \times\, 3!\, \times\, 7!\)

and so the probability that the three friends will sit next to each other is:

. . . . .\(\displaystyle \dfrac{8\, \times\, 3!\, \times\, 7!}{10!}\, =\, \dfrac{1}{15}\)

(b) The number of ways to choose two friends from the three friends is:

. . . . .\(\displaystyle \large{ \binom{3}{2} }\)

The number of outcomes for which the two chosen friends sit next to each other at one of the ends of the row and the third friend has a non-adjacent seat is:

. . . . .\(\displaystyle 2\, \times\, 7\, \times\, 2!\, \times\, 7!\)

while the number of outcomes for which the two chosen friends site next to each other but not at one of the ends of the row and the third friends has a non-adjacent seat is:

. . . . .\(\displaystyle 7\, \times\, 6\, \times\, 2!\, \times\, 7!\)

Hence the probability that exactly two of the three friends will sit next to each other is:

. . . . .\(\displaystyle \large{ \dfrac{\binom{3}{2}\, \left[\, 2\, \times\,7 \,\times\, 2! \,\times\, 7!\, +\, 7 \,\times\, 6 \,\times\, 2! \,\times\, 7!\, \right]}{10!}\, =\, \dfrac{7}{15} }\)



I don't understand how the answer to (b) was reached. I don't understand where any of the numbers come from in defining how many possibilities there are for each stated outcome. Could someone elaborate into how this was done?
Are you saying that you do understand the sample space and the numbers, etc, for the solution to part (a)?

Thank you! ;)
 
Question 7E-9 Three friends and seven other people are randomly seated in a row. Specify an appropriate sample space to answer the following two questions.
(b) What is the probability that exactly two of the three friends site next to each other?


I don't understand how the answer to (b) was reached. I don't understand where any of the numbers come from in defining how many possibilities there are for each stated outcome. Could someone elaborate into how this was done?
I will tell you how I do this question. Picture a row with 23 seats in it.
Seat the seven non-friends so that there are two empty seats on both sides of these people.
There are \(\displaystyle 7!\) ways to do that. Now there are eight groups of two-empty seats left.

There are three ways to select the two friends to sit together. There are two ways those two friends sit together.
Now there are eight places to put the two friends together and seven places left to place the single friend.
Realize that we are only concerned with the relative placements. NOT THE SEATS.

TOTAL: \(\displaystyle (7!)(3)(2)(8)(7)\).
 
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