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

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

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22 Nov 2006, 13:50
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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

Explanations pls.

Ans : B

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Director
Joined: 12 Jun 2006
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22 Nov 2006, 18:07
B or C for me

Driver's seat - 2 options (mom or dad)
Shotgun - 4 options (once mom or dad is in driver's seat)
Back seat behind driver - 2 options (girl 1 or girl 2)
Bench seat - 1 option
Back seat behind Shotgun (girl 1 or girl 2)

2(4)2(1)2 = 32

but what if one of the girls rode shottie; in that case wouldn't it be:
2(4)3(2)(1) = 48

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VP
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22 Nov 2006, 19:37
B. 32

step 1. on the driver seat, there are 2 possibilities.

step 2: consider the rest of the seats. Categorize them.

Category 1: the other parent on front seat.
back seat: daugthers have to separated. u have 2 choices.
so cat1: 2

Category 2: son at front.
back seat is similar to cat 1. 2
so cat 2: 2.

Category 3: one of the daughters is in the front seat. Remember either daughter could be in the front, so remember to multiply the back seat counts by 2 later.

Back seat: There are a parent, a son, and a daughter. They don't have any restrictions of seating, since a daughter is already seating in the front. so their arrangements is permutation of 3: i.e. 3!

So for Cat 3, there are totally 2*3! Multiply 3! by 2 is because the 2 daughters can exchange their seats.

Then total:

2*(2 + 2 + 2*3!) = 32.

Last edited by tennis_ball on 22 Nov 2006, 22:36, edited 2 times in total.

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Director
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22 Nov 2006, 20:33
Quote:
cat 3: 1 daughter in front. remeber either daughter could be at front.
back seat: permutation of 3: 3!.
so cat 3: 2*3!

tennis_ball,
will you explain the above a little more?

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VP
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22 Nov 2006, 22:35
ggarr wrote:
Quote:
cat 3: 1 daughter in front. remeber either daughter could be at front.
back seat: permutation of 3: 3!.
so cat 3: 2*3!

tennis_ball,
will you explain the above a little more?

OK. I edited my post.

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Manager
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23 Nov 2006, 19:04
tennis_ball wrote:
B. 32

step 1. on the driver seat, there are 2 possibilities.

step 2: consider the rest of the seats. Categorize them.

Category 1: the other parent on front seat.
back seat: daugthers have to separated. u have 2 choices.
so cat1: 2

Category 2: son at front.
back seat is similar to cat 1. 2
so cat 2: 2.

Category 3: one of the daughters is in the front seat. Remember either daughter could be in the front, so remember to multiply the back seat counts by 2 later.

Back seat: There are a parent, a son, and a daughter. They don't have any restrictions of seating, since a daughter is already seating in the front. so their arrangements is permutation of 3: i.e. 3!

So for Cat 3, there are totally 2*3! Multiply 3! by 2 is because the 2 daughters can exchange their seats.

Then total:

2*(2 + 2 + 2*3!) = 32.

Sorry but little confuse here.

Father=F
Mother=M
Son=S
Daughter1=D1
Daughter2=D2

Cat 1: F-M(D1-S-D2) --> times 2 since you can switch F and M = Total 2

Cat 2: F-S(D1-M-D2) --> times 2 since you can switch F and M = Total 2

Cat 3a: F-D1(3!) --> times 2 since you can switch D1 and D2 = Total 12
Cat 3b: M-D1(3!) --> times 2 since you can switch D1 and D2 = Total 12

Grand Total = 2+2+12+12 = 28
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Senior Manager
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23 Nov 2006, 19:46
Cat 1: F-M(D1-S-D2) --> times 2 since you can switch F and M = Total 2

Cat 2: F-S(D1-M-D2) --> times 2 since you can switch F and M = Total 2

d1 and d2 can flip places in both categories giving additional 4 arrangements.

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Senior Manager
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24 Nov 2006, 02:48
Three cases
Case 1 Front seats by parents and back seats by daughters and son thus we have 2 choices for front and 2 coices for back (as daughters need to separated) = 2*2=4
Case 2 Front seats occupied by a parent and son and back seats by daughters ,parent and son thus 2 possiblities for front seats and 2 for back Thus total =4

Case 3 front seats occupied by a daughter and a parent and back seats by son daughter and parent thus possibilites =2*2*3!=24

Total = Case1 + Case2 + Case3 =4+4+24 =32

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VP
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Location: Bangalore

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24 Nov 2006, 05:41
tennis_ball wrote:
B. 32

step 1. on the driver seat, there are 2 possibilities.

step 2: consider the rest of the seats. Categorize them.

Category 1: the other parent on front seat.
back seat: daugthers have to separated. u have 2 choices.
so cat1: 2

Category 2: son at front.
back seat is similar to cat 1. 2
so cat 2: 2.

Category 3: one of the daughters is in the front seat. Remember either daughter could be in the front, so remember to multiply the back seat counts by 2 later.

Back seat: There are a parent, a son, and a daughter. They don't have any restrictions of seating, since a daughter is already seating in the front. so their arrangements is permutation of 3: i.e. 3!

So for Cat 3, there are totally 2*3! Multiply 3! by 2 is because the 2 daughters can exchange their seats.

Then total:

2*(2 + 2 + 2*3!) = 32.

fantastic! You make permutations and combinations sound super easy!

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Manager
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24 Nov 2006, 10:55
Thanks Sumithra..

whoaaaa...
_________________

Who is John Galt?

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

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18 Mar 2017, 14:54
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Re: A family consisting of one mother, one father, two daughters   [#permalink] 18 Mar 2017, 14:54
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