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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 09 Mar 2008, 12:33
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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

This is an MGMAT CAT question, and the explanation the gave me is more convoluted than I want to learn right now. Can you think of an easier way? (I'll pose the official answer after a few responses)
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Re: Sitting in the Car... [#permalink] New post 09 Mar 2008, 14:03
32 ?


There are 2 possible drivers so answer is = 2 * ( N(parent front) + N(son front) + N (daughter front) )
N(parent front) = 2! (kid has to be in the middle)
N(son front) = 2! (parent has to be in the middle)
N(daughter front) = 2C1 * 3!

So result = 2 *(2+2+12) = 32
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Re: Sitting in the Car... [#permalink] New post 10 Mar 2008, 11:59
cumbersome but doable.

2c1 d1
3!

2c1 d2
3!

2c1 _
D1 _ D2


2C1 _
D2 _ D1

total = 32
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Re: Sitting in the Car... [#permalink] New post 10 Mar 2008, 12:19
Yes answer is 32, but I can't interpret what you did!
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Re: ways to sit in an Car [#permalink] New post 02 Aug 2008, 06:09
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MamtaKrishnia 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


all possible ways when one of the parents drive and no restrictions on how others can sit
= 2*4*3*2*1 = 48

answer has to be less than 48, eliminate C,D,E

take a 50-50 chance at A and B .... i'll take B :wink: ... just kidding lets solve it further

Possible ways when one of the parents in in drives seat and two daughter are together
drives seat = 2
co driver = 2 (either mom or son)
back seat = 2*2 ( 2 ways daughters can take left or right seats and 2 ways they can intechange)
total = 2*2*2*2 = 16

Our answer = 48-16 = 32 ... see i was right .... :)
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Re: ways to sit in an Car [#permalink] New post 05 Aug 2008, 01:27
MamtaKrishnia 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


Total number of ways = 2*4*3*2*1 = 48

However since two daughters cannot sit together which means we need to subtract the following number of combinations

Daughters sitting on seat 1&2, 2&3 on rear. = Number of combo in driver * front seat * first rear seat * middle seat * third rear seat = 2*2*2*2*1 = 16

Therefore possible arrangements = 48-16 = 32
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Re: ways to sit in an Car [#permalink] New post 05 Aug 2008, 08:54
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MamtaKrishnia 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


total seating arrangements = 2 4! = 48

total seating arrangement with two daughters sittng next to each other
= 2 ( two daughters sit rear left and middle seat) * 2 ( two daughters sit rear right and middle seat)* 2 (driver seat can be either parent ) * 2 ( front passenger seat .. either son or parent)
= 16

Ans = 40-16=32
B
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Re: Seating arrangements [#permalink] New post 21 Dec 2008, 11:36
Possible seating arrangements for driver seat = 2.

If the daughter occupies the front other seat then arrangements = 2.
And, the arrangements for the rear seat = 3*2

However, if the daughter does not occupy the front seat then arrangements = 2.
Arrangements for the rear seat - first seat can be taken in 2 ways (one of the two daughters).
Second seat can be taken in only one way (not the daughter).
Third seat can be taken in only one way.

Hence, total arrangements = 2*2*3*2 + 2*2*2
= 24 + 8 = 32

Possible seating arrangements for front second seat = 4.

Possible seating arrangements for rear seat = 3*2 = 6.

Hence, total seating arrangements = 2*4*3*2 = 48
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Re: Seating arrangements [#permalink] New post 22 Dec 2008, 10:56
krishan 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


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: Seating arrangements [#permalink] New post 23 Dec 2008, 23:59
# of times you can arrange family members 2x4x3x2=48
# of times you can arrange family members when two daughters sit next to each other 2x2x2x2=16
48-16= 32
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Re: Combination- Seating arrangement [#permalink] New post 19 Feb 2009, 12:09
shobuj40 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: Combination- Seating arrangement [#permalink] New post 20 Feb 2009, 13:56
shobuj40 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


Another way is adding up with daughter in front seat and no daughter in front seat

2.2.3! + 2.2.1.2

24+8 =32
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Re: Permutations [#permalink] New post 22 May 2010, 06:24
Expert's post
amitjash 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


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.
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Re: Permutations [#permalink] New post 05 Feb 2011, 04:11
thanks. i am rly bad on the probability questions... need to work on that asap.
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Re: Combinatorics: Family Seating [#permalink] New post 05 Apr 2011, 19:08
Hi

Please look here.

good-set-of-ps-85414.html

Regards,
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Re: Combinatorics: Family Seating [#permalink] New post 05 Apr 2011, 21:49
HelloKitty 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


Say I have my car seats:

[driver][front]
[back][back][back]

To keep the daughters apart is to either:
1) Have one daughter in front, OR
2) Have two daughters at the back but one seat apart

(possible drivers)x(possible daughter in front)x(arrangement of the remaining 3) = 2x2x3! = 24

How many ways to get (2):
(possible drivers)x(ways to arrange the daughter at the back)x(arrangement of the remaining 2) = 2x2!x2! = 8

8+24 = 32
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Re: Combination- Seating arrangement [#permalink] New post 06 May 2011, 06:53
In this problem, why do you assume that no daughters sit in the infront?

All we need to satisfy is "No daughters sit adjacent".
The problem does not say "no daughters in the front seat".

PS: I got this problem in MGMAT CAT and messed it up. :(
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Re: A family consisting of one mother, one father, two daughters [#permalink] New post 16 Aug 2014, 06:10
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Re: A family consisting of one mother, one father, two daughters   [#permalink] 16 Aug 2014, 06:10
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