Problem Solving

BTGmoderatorDC 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?

A. 28
B. 32
C. 48
D. 60
E. 120

OA B

Source: Manhattan GMAT

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 = [spoiler]32 = B[/spoiler]

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Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should ... 67256.html
- https://www.beatthegmat.com/counting-pr ... 44302.html
- https://www.beatthegmat.com/picking-a-5 ... 73110.html
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- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatori ... 73194.html
- https://www.beatthegmat.com/arabian-hor ... 50703.html
- https://www.beatthegmat.com/sub-sets-pr ... 73337.html
- https://www.beatthegmat.com/combinatori ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-se ... 71047.html
- https://www.beatthegmat.com/combinatori ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation ... 73915.html
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- https://www.beatthegmat.com/no-two-ladi ... 75661.html
- https://www.beatthegmat.com/laniera-s-c ... 15764.html

Cheers,
Brent

BTGmoderatorDC 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?

A. 28
B. 32
C. 48
D. 60
E. 120

We can analyze the problem using the following two cases: 1) the father is the driver and 2) the mother is the driver.

Case 1: The father is the driver.

If the father is the driver, we could have the following subcases:

i) The mother sits in the front row beside the father. Since the daughters refuse to sit next to each other, there are 2 seating arrangements in the back row: Dsd, dsD (D = elder daughter, d = younger daughter, and s = son).

ii) The son sits in the front row beside the father. Since the daughters refuse to sit next to each other, there are 2 seating arrangements in the back row: Dmd, dmD (m = mother).

iii) The elder daughter sits in the front row beside the father. Since, in this case, the two daughters will definitely not sit next to each other, there are 3! = 6 seating arrangements for the 3 remaining people who sit in the back row.

iv) The younger daughter sits in the front row beside the father. Like subcase (iii), there will be 6 seating arrangements in the back row.

As we can see from the above, if the father is the driver, there will be a total of 2 + 2 + 6 + 6 = 16 seating arrangements. We can make the identical argument when the mother is the driver. Thus, there will be an additional 16 seating arrangements, and, hence, we have a total of 16 + 16 = 32 seating arrangements.

Answer: B