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A family with 5 members (Father, Mother, two daughters and one son) pl

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A family with 5 members (Father, Mother, two daughters and one son) pl  [#permalink]

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New post 20 Apr 2017, 04:07
1
5
00:00
A
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D
E

Difficulty:

  65% (hard)

Question Stats:

60% (02:48) correct 40% (02:52) wrong based on 73 sessions

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A family with 5 members (Father, Mother, two daughters and one son) plan to go on a trip in car which has 5 seats (2 sets in front including driving seat and 3 seats at the back)
If all family members randomly sit on the the seats in car then what's the probability of an acceptable seating arrangement if two daughters do NOT want to sit next to each other and only one of the parents can drive???

Options:
A)1/120
B)1/60
C) 1/20
D) 4/15
E) 1/5

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Re: A family with 5 members (Father, Mother, two daughters and one son) pl  [#permalink]

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New post 20 Apr 2017, 04:18
3
let Driver seat be filled by Mother
front seat can be with 4 ways ... lets consider each
when Father at front .. 2D and 1 Son at back and 2 D cannot be together only 2 ways
when D1 is at front ... F D S 3! ways
when D2 at front ....... 3! ways
Now Son at front ... again 2 ways ....
so total 16 ways when mother as driver

so total 32 ways
32/120 = 4/15
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Re: A family with 5 members (Father, Mother, two daughters and one son) pl  [#permalink]

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New post 22 Mar 2019, 23:51
Case 1: both daughters at the back. seating arrangement: 2 seats in front, 3 in the back.

2 choices for driver, 2 choices for shotgun (one parent and son)

2 2
- -
2 1 1
- - -

total ways = 2*2*2*1*1 = 8

Case 2: one daughter in the front

2 2
- -
3 2 1
- - -

2*2*3*2*1 = 24 ways.

total ways to seat 5 people = 5! = 120.

probability = 32/120 = 4/15 (D)
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Re: A family with 5 members (Father, Mother, two daughters and one son) pl  [#permalink]

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New post 28 Apr 2019, 10:29
Saurabhminocha wrote:
Case 1: both daughters at the back. seating arrangement: 2 seats in front, 3 in the back.

2 choices for driver, 2 choices for shotgun (one parent and son)

2 2
- -
2 1 1
- - -

total ways = 2*2*2*1*1 = 8

Case 2: one daughter in the front

2 2
- -
3 2 1
- - -

2*2*3*2*1 = 24 ways.

total ways to seat 5 people = 5! = 120.

probability = 32/120 = 4/15 (D)



then what is meant by only one of the parents can drive?
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Re: A family with 5 members (Father, Mother, two daughters and one son) pl  [#permalink]

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New post 28 Apr 2019, 11:53
There are 2 cases possible
1. when father is driving-
2 sisters can sit in 4 ways ( when 1 in the front, other in the either of the 3 back seats and when both are in the window seats in the back), now both sister can be arranged in 2! ways and mother and son can be sit in 2! ways on remaining 2 seats
total number of ways possible in this case= 4*2!*2!=16
2. When mother is driving
total number of ways possible in this case= 4*2!*2!=16(we can find in the similar manner as the 1st case)
Total number of ways with the given restrictions=16+16=32
Probability= 32/120= 4/15
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Re: A family with 5 members (Father, Mother, two daughters and one son) pl   [#permalink] 28 Apr 2019, 11:53
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