Question

3 college friends brought their families for a reunion. John brought his wife, daughter and son, William brought his wife and son, and Jason brought his wife and daughter. Suppose they sat on a circular table randomly, what is the probability that the three friends will be sat next to one another?

Answer 1

4 + 3 + 3 = 10

Answer 2

that doesn't sound like a probability....

Answer 3

is answer
7!/10!

Answer 4

no idea

Answer 5

no answer for this?

Answer 6

no...im not an answering machine to be able to answer everything...

Answer 7

Lack of explanation ruins the solution process.

Answer 8

i would like to confirm answer before posting solution process.
answer is pretty weird.

Answer 9

i would assume the denominator would be 9!

Answer 10

both numerator and denominator has the circular symmetry of order 10, so both 10 would be dividing numerator and denominator ... and there isn't difference.
that's my guess

Answer 11

...that's impossible...

Answer 12

my guess would be (3!3!)/9!

Answer 13

i see ..
where i made mistake

Answer 14

wait.... i do too...

is it 3!/9!

Answer 15

my new guess would be
3!7!/10!

Answer 16

why 10! as denominator?

Answer 17

shouldn't it be 9! because (10-1)! = 9!

Answer 18

isn't there 10 people sitting on round table?

Answer 19

yes

Answer 20

circular permutation states that when there are n seats, the total ways of rearranging the n seats would be (n-1)!

Answer 21

so total perpumations will be
10!
since we consider circular symmetry, we should diving it by 10 ... since only 1 out of 10 is unique.

Answer 22

new sample space is 10!/10

now consider numerator.

Answer 23

....something soundss wrong there....

Answer 24

3 friends + 7 children and wives = 31 7!

Answer 25

why consider the arrangement of the 7 other people?

Answer 26

3 friends sitting together, you will have 3!7! permutations

Answer 27

since the only concern is the number of ways the friends can sit together...shouldn't it just be 3!

Answer 28

why not ... they are also included in sample space.

Answer 29

...do you consider the other things included in the sample space when doing probabilities?

Answer 30

isn't it just possible way of achieving something over total ways

Answer 31

yes ... as long as they satisfies condition.

Answer 32

I think:
\[\frac{ 3! 7! }{ 9! }\]

Answer 33

isn't it obvious 3!/9! <--- how small is this this?

Answer 34

Yes 3!7!/9!

Answer 35

consider circular permutation on numerator too.

Answer 36

i still don't know why 7! should be included

Answer 37

i did consider circular permutation in the numerator

Answer 38

\[\frac{(2 - 1)! 3!}{9!}\]

Answer 39

one out of ten is unique.

Answer 40

i considered two groups: one is the group of friends..the other group is the other 7

Answer 41

No it is (8-1)!3!/9!

Answer 42

why 8?

Answer 43

...on second thought...yes 8

Answer 44

I guess @siddhantsharan is going to explain

Answer 45

Look.

Total is clear 9!.

Now we consider the ways in which the three friends sit together.
Consider the three friends as a set.
Therefore you are left with 8 different entities.
Their permutaion is (8-1)!

Answer 46

i forgot each one of those 7 are unique

Answer 47

seems @siddhantsharan was two seconds late....good explanation though

Answer 48

Haha. Lol :)