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A family consisting of one mother, one father, two daughters
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Updated on: 27 Apr 2012, 04:07
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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
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Originally posted by pratikbais on 27 Apr 2012, 03:57.
Last edited by Bunuel on 27 Apr 2012, 04:07, edited 1 time in total.
Added the answer choices and the OA




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A family consisting of one mother, one father, two daughters
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27 Apr 2012, 04:07
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 Approach #1:Sisters can sit separately in two ways: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case: 2*4=8. Total=24+8=32. Approach #2:Total # of arrangements: Drivers seat: 2 (either mother or father); Front seat: 4 (any of 4 family members left); Back seat: 3! (arranging other 3 family members on the back seat); So. total # of arrangements is 2*4*3!=48. # of arrangements with sisters sitting together: Sisters can sit together only on the back seat either by the left window or by the right window  2, and either {S1,S2} or {S2,S1}  2 > 2*2=4; Drivers seat: 2 (either mother or father); Front seat: 2 (5  2 sisters on back seat  1 driver = 2); Back seat with sisters: 1 (the last family member left); So, # of arrangements with sisters sitting together is 4*2*2*1=16. 4816=32. Answer: B.
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Re: A family consisting of one mother, one father, two daughters
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28 Dec 2012, 20:05
pratikbais 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 Case 1: Both daughters in the back seat but seated separately Case 2: One daughter in front seat and the other in the middle of the back seat Case 3: One daughter in front seat and the other in the right back seat Case 4: One daugther in front seat and the other in the left back seat That's 4 positions and two daughters are interchangeable. Thus, 8. There are 2 ways to select a parent and 2 ways to seat the son and another parent. \(=4*2*2*2 = 32\) Answer: B




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Re: A family consisting of one mother, one father, two daughters
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27 Apr 2012, 10:30
pratikbais 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 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* (248) = 2*16 = 32. Thanks.



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Re: A family consisting of one mother, one father, two daughters
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26 Jul 2014, 10:23
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 D1D2 and D2D1, right window D1D2 and D2D1) *2 (we have to place the mother or father that is not driving)
Answer = 2*(4!4*2) = 32



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Re: A family consisting of one mother, one father, two daughters
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26 Jul 2014, 10:53
oss198 wrote: 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 D1D2 and D2D1, right window D1D2 and D2D1) *2 (we have to place the mother or father that is not driving)
Answer = 2*(4!4*2) = 32 _______________ Yes, that's correct.
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Re: A family consisting of one mother, one father, two daughters
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16 Sep 2014, 03:25
Bunuel 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
Approach #1:
Sisters can sit separately in two ways:
1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24
Or
2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case: 2*4=8.
Total=24+8=32.
Approach #2:
Total # of arrangements: Drivers seat: 2 (either mother or father); Front seat: 4 (any of 4 family members left); Back seat: 3! (arranging other 3 family members on the back seat); So. total # of arrangements is 2*4*3!=48.
# of arrangements with sisters sitting together: Sisters can sit together only on the back seat either by the left window or by the right window  2, and either {S1,S2} or {S2,S1}  2 > 2*2=4; Drivers seat: 2 (either mother or father); Front seat: 2 (5  2 sisters on back seat  1 driver = 2); Back seat with sisters: 1 (the last family member left); So, # of arrangements with sisters sitting together is 4*2*2*1=16.
4816=32.
Answer: B. Sorry Bunuel, but i didnt get the above in red. Why the 4* 2*2*1



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Re: A family consisting of one mother, one father, two daughters
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16 Sep 2014, 13:34
alphonsa wrote: Bunuel 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
Approach #1:
Sisters can sit separately in two ways:
1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24
Or
2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case: 2*4=8.
Total=24+8=32.
Approach #2:
Total # of arrangements: Drivers seat: 2 (either mother or father); Front seat: 4 (any of 4 family members left); Back seat: 3! (arranging other 3 family members on the back seat); So. total # of arrangements is 2*4*3!=48.
# of arrangements with sisters sitting together: Sisters can sit together only on the back seat either by the left window or by the right window  2, and either {S1,S2} or {S2,S1}  2 > 2*2=4; Drivers seat: 2 (either mother or father); Front seat: 2 (5  2 sisters on back seat  1 driver = 2); Back seat with sisters: 1 (the last family member left); So, # of arrangements with sisters sitting together is 4*2*2*1=16.
4816=32.
Answer: B. Sorry Bunuel, but i didnt get the above in red. Why the 4* 2*2*14 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.
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Re: A family consisting of one mother, one father, two daughters
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22 May 2017, 20:09
pratikbais 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 This problem can be tackled best via a direct method. There are two options for drivers (mother or father) = 2. Let's assume one of the sisters takes the passenger seat in the car. Then the number of spots for the rest of the family members is 3! = 3*2 2*3*2 = 12 ways * 2 sisters who could each sit in the front = 24 options. Now let's assume they both sit in the back seat. There are 2 options for the driver, 2 options for the passenger, and only one option for the back seat (the person separating the sisters) 2*2*1 = 4 * 2 sisters = 8 options.



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Re: A family consisting of one mother, one father, two daughters
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08 Sep 2017, 05:28
pratikbais 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 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 Posted from my mobile device
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Re: A family consisting of one mother, one father, two daughters
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11 Sep 2017, 15:56
pratikbais 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 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
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Re: A family consisting of one mother, one father, two daughters
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03 Feb 2018, 16:54
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 flipflop 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
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Re: A family consisting of one mother, one father, two daughters
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17 Aug 2018, 19:10
pratikbais 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 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 daughterdaughter 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 "daughterssittingadjacent" scenarios from the total number of "parentdriving" scenarios, we get: 48 – 16 = 32 The correct answer is B.



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Re: A family consisting of one mother, one father, two daughters
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18 Dec 2018, 23:18
There are two possibilities of arrangement. Case 1. When 1 daughter is in front. Total ways = 2(no of ways to choose driver) x 2(no of ways to choose 1 daughter for front seat) x 3!(arrangement of 3 people in back seats. = 2 x 2 x 6 = 24 ways. Case 2. When both daughters are sitting in the back and one person is sitting between them. Total ways = 2(ways to choose driver) x 2( ways to choose person to be seated in front) x 2(arrangement of two daughters in the left and right of person sitting in the back seat)= 8 ways. So, total no of arrangements = 32. Hence, b is the answer.



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Re: A family consisting of one mother, one father, two daughters
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23 Nov 2019, 10:19
pratikbais 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 considering parent taking driver seat = 2c1 and daughter front seat = 1 last 3 seats can be occupied in 3! ways case 2: 3 seats occupied by 2 daughters ; 2c1 and 1 seat among son and 1 parent ; 2c1 total ways ; 2*(2c1*3!+2c1*2c1 ) = 32 IMO B




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