enfinity wrote:

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

OR

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.