Thanks man! Here it comes:
You have four friends: Jim, Arun, Soma, and Eric. How many ways can you rearrange the individuals in a row so that Soma and Eric do not sit next to each other.
My approach (and the one presented in the book i've mentioned):
Find all possibilities: 4!=24 (certain scenarios are not possible; therefore, we need to subtract these impossibilities):
Find the ways in which S and E sit next to each other:
E S _ _
_ E S _
_ _ E S
S E _ _
_ S E _
_ _ S E
Therefore, 24 - 6 = 18
This approach is similar to the other one, but it has a mistake.
1st. All possibilities = 4! = 24
2nd. Take ES as 1 person. So now we have 3 people: J, A & ES --> 3! = 6
3nd. Finally, it is necessary to account for different arrangements within ES --> 2! = 2
=> All - P (they DO sit together) = 24 - 3! * 2! = 24 - 12 = 12
By multiplying by 3 instead of 3! you are not accounting for the different positions of J and A. Have a look at the following:
S E J A
J S E A
J A S E
Is not the same as
S E A J
A S E J
A J S E
But the approach presented on that book does not take that into account.
Back to the original problem:
=> All - P (they DO sit together) = 6! - 5! * 2! = 720 - 120 * 2 = 480
Hope that helps.
Also, please know that probability is not my forte.
Please kudos if my post helps.