Clarification

knight247 wrote:In a certain bus, seats are arranged such that there are 4 in each row: 2 are to the left of the central aisle and 2 are to the right of the central aisle. Seats are considered to be "next to one another" only if they are in the same row and on the same side of the aisle. If the bus has 10 rows of seats, what is the probability that two randomly chosen seats are in the same row but are NOT "next to one another"?

OA is [spoiler]2/39[/spoiler]

Here is how i solved it:

Total number of possibilities is 40C2=780

Now to find two seats in the same row, we first pick any one out of the 10 rows in 10C1 Ways
Now in this row, we pick any one seat in 4C1 ways.
Now, the next seat we pick has to be on the other side of the aisle, so we pick the next one in 2C1 ways.

Altogether we have 10C1*4C1*2C1= 80

80/780=8/78=4/39

I'm getting an additional 2 somewhere. Hoping to get a clarification. Thanks

It might be easier to solve this one using probability (vs. counting)

P(same row but not adjacent) = P (select any seat first AND then select seat in same row but not adjacent)
= P (select any seat first) x P(select seat in same row but not adjacent)
= 1 x [spoiler]2/39[/spoiler]
= [spoiler]2/39[/spoiler]

Aside: Once we have selected the first seat, there are 39 seats remaining. Of those 39 seats, 2 are in the same row, but not adjacent to the first seat.

Cheers,
Bremt

knight247 wrote:In a certain bus, seats are arranged such that there are 4 in each row: 2 are to the left of the central aisle and 2 are to the right of the central aisle. Seats are considered to be "next to one another" only if they are in the same row and on the same side of the aisle. If the bus has 10 rows of seats, what is the probability that two randomly chosen seats are in the same row but are NOT "next to one another"?

OA is [spoiler]2/39[/spoiler]

Here is how i solved it:

Total number of possibilities is 40C2=780

Now to find two seats in the same row, we first pick any one out of the 10 rows in 10C1 Ways
Now in this row, we pick any one seat in 4C1 ways.
Now, the next seat we pick has to be on the other side of the aisle, so we pick the next one in 2C1 ways.

Altogether we have 10C1*4C1*2C1= 80

80/780=8/78=4/39

I'm getting an additional 2 somewhere. Hoping to get a clarification. Thanks

The problem with these calculations is in bolded red font above.
Even though you are using combinations, you are making the order matter (even though it shouldn't)

For example, if we have have seats 1, 2, 3, and 4 in a row. And we have two people: A and B, your calculations are treating the following cases as separate cases:
case 1) first select A for seat 1, and then select B for seat 3
case 2) first select B for seat 3, and then select A for seat 1
These two identical cases have been counted separately in your solution.

So, if we divide 80 by 2 (to account for duplication), we get 40/780 = 2/39

Cheers,
Brent