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

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A family consisting of one mother, one father, two daughters [#permalink]

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27 Apr 2012, 03:57
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Apr 2012, 04:07, edited 1 time in total.

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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

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.

48-16=32.

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Re: A family consisting of one mother, one father, two daughters [#permalink]

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27 Apr 2012, 10:30
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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* (24-8) = 2*16 = 32.

Thanks.

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Re: A family consisting of one mother, one father, two daughters [#permalink]

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28 Dec 2012, 20:05
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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.

$$=4*2*2*2 = 32$$

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Re: A family consisting of one mother, one father, two daughters [#permalink]

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26 Jul 2014, 10:23
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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)

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Re: A family consisting of one mother, one father, two daughters [#permalink]

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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 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)

_______________
Yes, that's correct.
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Re: A family consisting of one mother, one father, two daughters [#permalink]

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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.

48-16=32.

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 [#permalink]

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16 Sep 2014, 13:34
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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.

48-16=32.

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.
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Re: A family consisting of one mother, one father, two daughters [#permalink]

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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 [#permalink]

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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

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Re: A family consisting of one mother, one father, two daughters [#permalink]

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11 Sep 2017, 15:56
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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.

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Re: A family consisting of one mother, one father, two daughters   [#permalink] 11 Sep 2017, 15:56
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