Problem Solving

sparkle6 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

[spoiler]Answer: B. I narrowed it down to A & B but could not figure out how to solve for the exact number.[/spoiler]

Knight247's approach is great. Here's another approach.

Number the seats as follows:
Seat #1: driver's seat
Seat #2: passenger's seat
Seats #3, 4, 5 back seats

First ignore rule about the sisters not sitting together.
Stage 1: seat a parent in seat #1. We can accomplish this in 2 ways
Stage 2: seat someone in seat #2. We can accomplish this in 4 ways.
Stage 3: seat someone in seat #3. We can accomplish this in 3 ways.
Stage 4: seat someone in seat #4. We can accomplish this in 2 ways.
Stage 5: seat someone in seat #5. We can accomplish this in 1 way.
Total number of arrangements (ignoring the "sisters together rule") = 2x4x3x2x1=48

At this point we need to subtract from 48 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
Stage 1: seat a parent in seat #1. We can accomplish this in 2 ways
Stage 2: seat a sister in seat #3 We can accomplish this in 2 ways.
Stage 3: seat the other sister in seat #4. We can accomplish this in 1 way.
Stage 4: seat someone in seat #2. We can accomplish this 2 ways.
Stage 5: seat someone in seat #5. We can accomplish this 1 way.
Total number of arrangements = 2x2x1x2x1 = 8

case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get 8 arrangements


So the final answer is 48 - 8 - 8 = 32

Cheers,
Brent

Kemmy G wrote:I thought I understood permutation and combination, Brent, but you just lost me with this! Could you please explain it again, step by step? I tried to work it with the slot method and with the anagram table method, even with the nPr method, and I was still nowhere near 32. I sorta understand how you got the answer. What I don't understand is why you worked it that way. Please explain again!

No problem.

I'll begin by saying that I'm not a big fan of permutations. The truth of the matter is that there are very few true permutation questions on the GMAT. In fact, the majority of questions (including all true permutation questions) can be solved using the Fundamental Counting Principle (FCP). In my learning materials, I never use the word "permutations." You'll also find that the word "permutation" appears only once in the OG12, and they don't bother supplying the permutation formula.

To apply the FCP, you need to take a required task (here the task is to seat 5 people) and break it into stages. Now the restriction about the sisters 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 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
First break the task into stages:
Stage 1: seat a parent in seat #1.
Stage 2: seat someone in seat #2.
Stage 3: seat someone in seat #3.
Stage 4: seat someone in seat #4.
Stage 5: seat someone in seat #5.

Now we determine the number of ways to accomplish each stage.
Stage 1: Only a parent can sit here. So, this stage can be accomplished in 2 ways.
Stage 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: 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: this stage can be accomplished in 2 ways.
Stage 5: this stage can be accomplished in 1 way

By the FCP, the total number of ways to accomplish all 5 stages (and seat every family member) = 2x4x3x2x1=48

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.
Stage 2: seat a sister in seat #3
Stage 3: seat the other sister in seat #4.
Stage 4: seat someone in seat #2.
Stage 5: seat someone in seat #5.
Once we have completed all 5 stages, we will have seated every family member BUT this time we are ensuring that the sisters DO sit together.

Stage 1: must be 1 of 2 parents. So, this stage can be accomplished in 2 ways.
Stage 2: must be 1 of 2 sisters. So, this stage can be accomplished in 2 ways.
Stage 3: once we have seated a sister in seat #3, only 1 sister remains. So, this stage can be accomplished in 1 way.
Stage 4: at this point, we have seated 3 people, so only 2 people remain. So, this stage can be accomplished in 2 ways.
Stage 5: one person remaining. So, this stage can be accomplished in 1 way.

Total number of arrangements where the sisters are in seats #3 and #4 = 2x2x1x2x1 = 8

case 2: the sisters are in seats #4 and #5
We can follow the same steps as above to get 8 arrangements

So the final answer is 48 - 8 - 8 = 32

Cheers,
Brent

Kemmy G wrote:Thanks Brent!

I understand it much better with the breakdown, but I'm still going to have to go over it with a fine tooth comb to fully internalise and understand it. Thanks very much!

To help you solidify this technique, it might help to review other counting questions solved using the FCP. Here are two such questions:

https://www.beatthegmat.com/3-rings-t92252.html
https://www.beatthegmat.com/combinatorics-04-t91876.html

You can also watch a free video lesson on the FCP at: https://www.gmatprepnow.com/module/gmat-counting (it's lesson #3)

Cheers,
Brent