cagheo wrote:A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
1) 28
2) 32
3) 48
4) 60
5) 120
Another, slightly longer, approach:
The restriction about the sisters is somewhat problematic, so let's IGNORE the rule and seat all 5 people
without obeying that restriction.
Then once I determine the total number of arrangements, I'll subtract the number of arrangements where the
sisters are sitting together.
Number the seats as follows:
Seat #1: driver's seat
Seat #2: passenger's seat
Seats #3, 4, 5: back seats
# of arrangements where we ignore rule about the sisters not sitting together
Take the task of seating all 5 people and break into stages.
Stage 1: Seat someone in seat #1
Only a parent can sit here. So, this stage can be accomplished in
2 ways.
Stage 2: Seat someone in seat #2
Once we have seated someone in seat #1, there are 4 people remaining. So, this stage can be accomplished in
4 ways.
Stage 3: Seat someone in seat #3
At this point, we have already seated 2 people, so there are now 3 people remaining.
So, this stage can be accomplished in
3 ways.
Stage 4: Seat someone in seat #4
There are 2 people remaining, so this stage can be accomplished in
2 ways.
Stage 5: Seat someone in seat $5
This stage can be accomplished in
1 way
By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat all 5 people) in
(2)(4)(3)(2)(1) ways
48 ways
So there are
48 different ways to seat the family such that a parent drives. At this point, the
48 different arrangements
include arrangements where the sisters are seated together. So, we need to SUBTRACT the number of arrangements where the sisters are seated together.
There are two cases where the sisters are together.
case 1: the sisters are in seats #3 and #4
case 2: the sisters are in seats #4 and #5
case 1: the sisters are in seats #3 and #4
Once again, we'll take the task of seating everyone and break it into stages:
Stage 1: seat a parent in seat #1.
Must be 1 of 2 parents. So, this stage can be accomplished in
2 ways.
Stage 2: seat a sister in seat #3
Must be 1 of 2 sisters. So, this stage can be accomplished in 2 ways.
Stage 3: seat the other sister in seat #4.
Once we have seated a sister in seat #3, only 1 sister remains. So, this stage can be accomplished in
1 way.
Stage 4: seat someone in seat #2.
At this point, we have seated 3 people, so only 2 people remain. So, this stage can be accomplished in
2 ways.
Stage 5: seat someone in seat #5.
One person remaining. So, this stage can be accomplished in
1 way.
By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus seat the sisters are in seats #3 and #4) in
(2)(2)(1)(2)(1) ways
8 ways
case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get
8 more arrangements
So the final answer is
48 -
8 -
8 =
32
Aside: For more information about the FCP, watch our free video:
https://www.gmatprepnow.com/module/gmat-counting?id=775
Cheers,
Brent
