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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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09 Mar 2008, 13:33
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Difficulty:

65% (hard)

Question Stats:

51% (02:52) correct 49% (01:56) wrong based on 74 sessions

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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)
[Reveal] Spoiler: OA
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Re: Sitting in the Car... [#permalink]

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09 Mar 2008, 15:03
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]

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10 Mar 2008, 12:59
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]

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10 Mar 2008, 13:19
Yes answer is 32, but I can't interpret what you did!
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Re: ways to sit in an Car [#permalink]

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02 Aug 2008, 07:09
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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 ... 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]

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05 Aug 2008, 02:27
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]

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05 Aug 2008, 09:54
1
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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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21 Dec 2008, 12:36
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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22 Dec 2008, 11:56
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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24 Dec 2008, 00:59
# 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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19 Feb 2009, 13:09
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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20 Feb 2009, 14:56
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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22 May 2010, 07:24
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.

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05 Feb 2011, 05:11
thanks. i am rly bad on the probability questions... need to work on that asap.
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05 Apr 2011, 20:08
Hi

good-set-of-ps-85414.html

Regards,
Subhash
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05 Apr 2011, 22:49
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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06 May 2011, 07:53
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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]

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

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28 Dec 2015, 16:10
solved it this way:

since 2 parents can be in the driving seat, we have 2C1, or 2 ways.
we then have 4C1*3C1*2C1*1C1 ways we can arrange the rest of the seats. in total: 2*4*3*2*1=48 ways. This is the total number of ways we can arrange everyone, but not respecting the condition. We can eliminate D and E right away.

now, the 2 girls do not have to sit together. Let's see how many ways, in which 2 girls sit together, are. Suppose 2 girls is 1 person.
we can do this in 2C1 (parent driving) * 3C1 (front seat) * 2C1 * 1C1. we have 2*3*2*1 or 12 ways. if 2 sisters are in the back.
and 2C1*1*2C2*1C1=4 ways in which 1 sister is in front. 12+4=16.

now, there must be 48-16 =32 ways in which the 2 girls do not sit together.
A family consisting of one mother, one father, two daughters   [#permalink] 28 Dec 2015, 16:10
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