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# In a railway compartment, there are 2 rows of seats facing

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In a railway compartment, there are 2 rows of seats facing  [#permalink]

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15 Apr 2014, 03:15
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Difficulty:

65% (hard)

Question Stats:

60% (02:23) correct 40% (02:14) wrong based on 175 sessions

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In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?

A.172000

B.12600

C.45920

D.43200
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Re: In a railway compartment, there are 2 rows of seats facing  [#permalink]

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25 Oct 2015, 22:13
6
3
suryav wrote:
In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?

A.172000

B.12600

C.45920

D.43200

Here is another way to think about it - each seat is distinct. 5 face forward and 5 face back.
Start with the 4 people who want to face forward - for the first one, there are 5 options, for next one, 4 options and so on till we get 5*4*3*2.
Next, go on to 3 people who want to face back - for the first one, there are 5 options, for the next one, 4 options and so on till we get 5*4*3.
Now, we have 3 seats leftover and 3 indifferent people who can be arranged in 3*2*1 ways.
All in all, we can arrange 10 people in 5*4*3*2*5*4*3*3*2*1 = 14400 * 3 = 43200

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##### General Discussion
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Posts: 47983
Re: In a railway compartment, there are 2 rows of seats facing  [#permalink]

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15 Apr 2014, 03:26
2
5
suryav wrote:
In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?

A.172000

B.12600

C.45920

D.43200

A --> E
B --> F
C --> G
D --> X
X --> X

The total number of arrangements = $$5!*5!*C^1_3=43,200$$, where 5! is the number of ways to arrange 5 people facing forward, 5! is the number of ways to arrange 5 people facing backward, and $$C^1_3$$ is the number of ways to choose 1 out of 3 indifferent who will face forward (the remaining 2 will face backward).

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In a railway compartment, there are 2 rows of seats facing  [#permalink]

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25 Oct 2015, 08:50
I can understand the last 3! (for the 3 who are indifferent) but should not we do factorial for 4 facing front and 3 facing rear?

How do we get 5! * 5!?
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In a railway compartment, there are 2 rows of seats facing  [#permalink]

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25 Oct 2015, 08:58
dina98 wrote:
I can understand the last 3! (for the 3 who are indifferent) but should not we do factorial for 4 facing front and 3 facing rear?

How do we get 5! * 5!?

In each row the number of arrangements is 5! because there are 5 seats. Hence, 5!*5!.

Next, there are 3 people which does not care where they sit. $$C^1_3=3$$ is the number of ways to choose which one of them will sit with 4 wishing to sit facing forward and the remaining 2 will automatically be placed with 3 wishing to sit facing towards the rear.
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In a railway compartment, there are 2 rows of seats facing  [#permalink]

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26 Oct 2015, 01:30
This is permutation since order needs to be considered
4 persons have a choice of 5 seats (5P4), again 3 persons have a choice of 5 seats (5P3)and the remaining 3 persons in the remaining 3 seats(3!).
Total number of arrangements is 5P4*5P3*3!=43,200
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Re: In a railway compartment, there are 2 rows of seats facing  [#permalink]

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15 Oct 2017, 03:00
Can someone help me solve this in slot method. I tried but was not successful.

I considered FFFFRRRIII , F being facing forward, R being facing rear and I being indifferent. this can be arranged in 10!/4!3!3!. What is wrong in my approach.

Thanks.
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Re: In a railway compartment, there are 2 rows of seats facing  [#permalink]

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21 Oct 2017, 02:09
1
Hi,
I think you considered all in one row. But question clearly states - there are 2 rows of seats facing each other with accommodation for 5 in each.

F=>F1 F2 F3 F4 I1
R=>R1 R2 R3 I2 I3

Lets select and arrange F row for (F1 to F4)- > 5C4*4! .= 5*4! = 5! = 120
Lets select and arrange R row for (R1 to R3)- > 5C3*3!.=10 * 6 = 60
Now remaining 3 indifferent can be arranged in 3! = 6 ways.

Total = 120*60*6 = 43200

Ans : - D
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Re: In a railway compartment, there are 2 rows of seats facing &nbs [#permalink] 21 Oct 2017, 02:09
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