A family consisting of one mother, one father, two daughters

Steps to solve this problem -
People - 1M, 1F, 1S, 2D (D1,D2)
1. Choose a parent to drive the sedan.
2. Find the ways to seat the daughters.
3. Place the remaining 3 family members.
4. Finally multiply results from steps 1, 2 and 3.

1. Ways to choose a parent to drive = 2 (One person seated, total remaining = 4).
2. Arrangement in which daughters sit separate = Total Arrangements of 4 people - Arrangements with D1 and D2 glued together.

-> 4! - 4*2*1

4*2*1 - the pair of daughters can take 2 out of 3 consecutive spots at the back seat. Also they can interchange seats D1D2 and D2D1 are different arrangements. So total 4.
Remaining two people sit in 2*1 ways.

(This takes care of step 3 as well)

Finally step 4 - Total = 2* (4! - 4*2*1 ) = 2* (24-8) = 2*16 = 32.

Thanks.

Could someone tell me is my approach is correct?

1) Total of arrangement in the driver seat : 2 (mother or father)

2) Total arrangement of the four others : 4!

3) Total ways with the 2 daughters next to each other : 4 (left window D1-D2 and D2-D1, right window D1-D2 and D2-D1) *2 (we have to place the mother or father that is not driving)

Answer = 2*(4!-4*2) = 32

_______________
Yes, that's correct.

Sorry Bunuel, but i didnt get the above in red.
Why the 4*2*2*1

4 ways to sit sisters together;
2 ways to fill drivers seat (mother or father);
2 ways to fill front seat (3 people are already distributed, so 2 are left);
1 way to fill the remaining back seat.

Total = 4*2*2*1.

Hope it's clear.

2C1- Parent for driving seat

Case 1: one daughter sits on front seat
2C1- to select front seat daughter
3! - arrangement of rest 3 on back seats

Case 2: Both sisters sit on back seat at extreme corners
2! - arrangement of sisters
2! - Arrangement of one parent and the boy


total outcomes = 2C1*(2C1*3!+2!*2!)
= 2*(12+4) = 32

Answer Option B

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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 same argument when the mother is the driver. Thus, there will be another 16 seating arrangements, and hence we have a total of 16 + 16 = 32 seating arrangements.

Answer: B

Hi All,

These types of questions can be approached a couple of different ways. There's a "visual" aspect to this question that can help you to take advantage of some shortcuts built into the prompt, so I'm going to use a bit of "brute force" and some pictures to answer this question. Since we're arranging people in seats, we'll end up doing some "permutation math."

M = Mother
F = Father
D1 = 1st Daughter
D2 = 2nd Daughter
S = Son

Front Back
_ _ _ _ _
1st spot = driver

We have 2 restrictions that we have to follow:
1) Either the Father or Mother must be the driver
2) The two daughters CANNOT sit next to one another

Let's put the Mother in the driver's seat and count up the possibilities:

M F (2)(1)(1) Here, the two daughters have to be separated by the son, but either daughter could be in the "first back seat" = 2 options

M D1 (3)(2)(1) Here, with the first daughter up front, the remaining 3 people (F, D2 and S) can be in any of the back seats = 6 options
M D2 (3)(2)(1) Here, we have the same situation, but with the second daughter up front… = 6 options
M S (2)(1)(1) Here, with the son up front, we have the same scenario as we had when the Father was up front = 2 options

Total options with Mother driving = 2+6+6+2 = 16 options

Now we can take advantage of the shortcut I mentioned earlier - We can flip-flop the Mother and Father in the above examples. This will gives us another 16 options with the Father driving.

Total options: 16 + 16 = 32 options

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Official Solution (Credit: Manhattan Prep)

The easiest way to solve this question is to consider the restrictions separately. Let’s start by considering the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters won't sit next to each other.

This means that…
2 people (mother or father) could sit in the driver’s seat
4 people (remaining parent or one of the children) could sit in the front passenger seat
3 people could sit in the first back seat
2 people could sit in the second back seat
1 person could sit in the remaining back seat

The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3 × 2 × 1 = 48

We must subtract from these 48 possible seating arrangements the number of seating arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each other is if they are both sitting in the back.

This means that…
2 people (mother or father) could sit in the driver’s seat
2 people (remaining parent or son) could sit in the front passenger seat

Now for the back three seats we will do something a little different. The back three seats must contain the two daughters and the remaining person (son or parent). To find out the number of arrangements in which the daughters are sitting adjacent, let’s consider the two daughters as one unit. The remaining person (son or parent) is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill.
There are 2 × 1 = 2 ways to seat these two units.
However, the daughter-daughter unit could be d1d2 or d2d1
We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the back.
We could also have manually counted these possibilities:
d1d2X, d2d1X, Xd1d2, Xd2d1

Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier:
(2 × 2) × 4 = 16
front back

If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving" scenarios, we get: 48 – 16 = 32

The correct answer is B.

Bunuel, i was doing 5!-16
why are we considering the one condition as driver to be either mother/father while taking total arrangement- arrangement of sister sitting together

Hi kunalc20,

If you want to start with 5! (re: the assumption that any person could sit in any seat), and then 'subtract out' all of the possibilities that DON'T fit the various 'seating rules', then that's fine, but you would have to remove far more than just 16 of the 120 possibilities. ONLY the Mother or Father can sit in the driver's seat, so that alone removes (3)(4!) = 72 of the options (these are the options in which one of the three CHILDREN is sitting in the driver's seat). Once you've dealt with that issue - and you have 48 options remaining - you then also have to remove the options in which the Mother or Father is driving AND the 2 sisters are sitting next to one another in the back seat.

GMAT assassins aren't born, they're made,
Rich

Sorry I am probably a bit dumb on this - can someone explain my mistake

if I do a manual ways I got the following

1. m(driving) f(front) - 2 daughters 1 son = 4
2. f(driving) m(front) = 2 daughters 1 son = 4
3. m (driving) s(front) = 2 daughters 1 father = 4
4. f (driving) s (front) - 2 daughters 1 mother = 4
5. m (driving) d1 front = d2/s/f back = 6
6. f (driving) d1 front = d2/s/m back =6
7. f (driving) d2 front = d1/s/m back =6
8. m (driving) d2 front = d1/s/f back = 6

so 40 - where is that missing 8?

Hi levin343,

The prompt tells us that the two daughters refuse to sit next to one another, so when you put both daughters in the back seat, you have be careful that none of the options that you're calculating place the daughters side-by-side.

For example, with 2 daughters and the son in the back, there are only two possible arrangements...

D1/S/D2
D2/S/D1

...not the four arrangements that you have in your explanation.

GMAT assassins aren't born, they're made,
Rich

This is how I reasoned the problem. Notice, that if a daughter sits up front with a parent, the daughters are not considered to be sitting next to each other, therefore we can split this problem into one of two cases. Does a daughter sit up front with a parent or not? We can then add the cases together

Case 1 a daughter sits up front with a parent,. we have 2 options for the driving parent * 2 options for a daughter in passenger seat *3*2*1 for the back seat so we get 2*2*3*2*1 = 24 for this case

Case 2 a daughter does not sit up front with a parent. Then we have 2 options for the parent driving * 2 (the other parent or son for the passenger) * (Now for the backseat 3 people remain but the daughters cant sit together so we can treat them like one unit so we have 2*1 choices for the back seat). so we get get 2*2*2=8 options for this case

24+8 = 32 possible seating arrangements.

MartyMurray KarishmaB let me know if this looks correct.

Taking all mother as driver scenarios. For father as driver scenarios, both will be multiplied by 2.

Scenario 1:
M F
D1 S D2

For D1, D2 arrangement in the back seats (2)
For F, S arrangement between back and front seats (2)
So, 2 * 2 = 4

Multiplying by 2 again, to consider father. Therefore this gives me 8.

Scenario 2:
M D1
D2 S F

For D1, D2 front and back seat exchange (2)
For F and S seating arrangement in the backseats (2)
A D1 or D2 can be seated in the backseats is (3) ways
So, 2 * 2 * 3 = 12

Multiplying by 2 again, to consider father. Therefore this gives me 24.

Adding both scenarios above = 24+8 = 32 (B)

Looks fine though I would worry if I missed some cases in solving it this way. I would much rather go the 'All - NOT Allowed' way.

I pick a driver in 2 ways.

All other 4 people can be arranged in 4! ways.

Not Allowed Cases - Make the daughters sit together (make a group) in the back seat. Pick one more person for the backseat (out of father and son) in 2 ways.
Now arrange the group and the third person in 2! * 2! (arrangement of the two daughters)
Total arrangements = 2 * 2 * 2 = 8

Hence Allowed arrangements = 2 * (4! - 8) = 32

Answer (B)

Check out this video on Permutations:

Thank you, Karishma. I agree I can missing a case or couple. All - Total # of arrangements where daughters would sit together would be lesser error proof.