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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] New post 01 Sep 2008, 21:30
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A
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E

Difficulty:

  45% (medium)

Question Stats:

58% (02:02) correct 42% (01:53) wrong based on 42 sessions
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-family-consisting-of-one-mother-one-father-two-daughters-131430.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Sep 2013, 08:27, edited 1 time in total.
Added the OA.
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Re: A family consisting of one mother, one father, two daughters [#permalink] New post 19 Sep 2013, 08:28
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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.

Answer: B.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-family-consisting-of-one-mother-one-father-two-daughters-131430.html
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Re: PS - combinatroics [#permalink] New post 01 Sep 2008, 21:46
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gmatkudi 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?
28
32
48
60
120

IMO 32

Scenario 1:
Mom driving and dad in front.
2 seating arrangements

Scenario 2:
Mom driving and son in front
2 seating arrangements possible

Scenario 3:
Mom driving and one of the sisters in the front.
Sister 2 has (3*2*1 = 6) possible arrangements to sit.

Scenario 4:
Mom driving and the other sister in front'
Same logic as scenario 3....6 possible arrangements

Total scenarios for mom driving = 2+2+6+6 =16
Same thing with Dad driving.
Hence total possible scenarios = 16*2 = 32
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Re: PS - combinatroics [#permalink] New post 01 Sep 2008, 22:21
yes, i also got the same thing..
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Re: PS - combinatroics [#permalink] New post 01 Sep 2008, 22:26
gmatkudi 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?
28
32
48
60
120


Total # ways possible (with only one constraint of one parent sitting in front)
= 4! *2 = 48

# ways, when both the daughters are sitting together: (P1-Parent1, P2- Parent2, S-Son, D- Daughter)
P1 S D D P2 -> 2 ways (for daughter) + 2 ways (when P1 is swapped with P2)= 4 ways
P1 S P2 D D -> 2 ways (for daughter) + 2 ways (when P1 is swapped with P2)= 4 ways
P1 P2 D D S -> 2 ways (for daughter) + 2 ways (when P1 is swapped with P2)= 4 ways
P1 P2 S D D -> 2 ways (for daughter) + 2 ways (when P1 is swapped with P2)= 4 ways
Total = 16 ways.

Number of ways both the daughters NOT sit near each other = 48-16= 32

WHat is OA?
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Re: PS - combinatroics [#permalink] New post 02 Sep 2008, 11:20
gmatkudi 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?
28
32
48
60
120


PX
DDX

= Total No of ways - two daughters sit together
= 2*4! - 2 (choose 1 Parent for Driver seat)* 2(Daughters can be arranged themselves) * 2 (DD and X (backseat) arranged in themselves) * 2( X and X can be aranged in two ways)
=48 -2*2*2*2
=32
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Re: A family consisting of one mother, one father, two daughters [#permalink] New post 19 Sep 2013, 06:44
I tried a different solution but was unsuccessfull.
Can anyone see where I went wrong?

I tried it so:
Number of all possible seating arrangements (2*4!) = 48
Number of seating arrangements where girls seat together and one parent drives:
2*2*3!.....

and then subtracting the two.

Can anyone see where I went wrong?
Re: A family consisting of one mother, one father, two daughters   [#permalink] 19 Sep 2013, 06:44
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