seyna wrote:A family with 1 mother, 1 father, and 3 children sit in a car with 2 front seats and 3 back seats. Two of the 3 children (Matt and Peter) cannot sit next to each other as they fight too often. If only the mother and father can drive and sit in the driver's seat, how many seating arrangements can be made?
The restriction about the boys (Matt and Peter) is somewhat problematic, so I decided to ignore the rule and seat all 5 people
without obeying that restriction.
Then once I determine the total number of arrangements, I subtract the number of arrangements where the
boys 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 Matt and Peter 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 Matt and Peter are seated together. So, we need to SUBTRACT the number of arrangements where Matt and Peter are seated together.
There are two cases where the boys are together.
case 1: Matt and Peter are in seats #3 and #4
case 2: Matt and Peter are in seats #4 and #5
case 1: Matt and Peter 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 either Matt or Peter in seat #3
Must be 1 of 2 boys. So, this stage can be accomplished in 2 ways.
Stage 3: seat the other boy in seat #4.
Once we have seated a boy in seat #3, only 1 boy 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 two boys in seats #3 and #4) in
(2)(2)(1)(2)(1) ways
8 ways
case 2: the boys 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
Cheers,
Brent
By the way, this question is nearly identical to this one:
https://www.beatthegmat.com/a-family-con ... 03048.html
