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There are 1111 cars in the parking lot each with number 1 to

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There are 1111 cars in the parking lot each with number 1 to [#permalink] New post 13 Jun 2006, 21:55
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There are 1111 cars in the parking lot each with number 1 to 1111 and no two cars have the same number. At 5pm when they all leave the parking lot one by one, what would be the probability that numbers of the first four cars to leave are in increasing order?



seems laxi is in short break for a while. everybody is appreciated for his/her inputs.
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 [#permalink] New post 13 Jun 2006, 23:02
Ok I know

the first car can have numbers 1 - 1108
The second car can have numbers 2 - 1109
The third car can have numbers 3 - 1110
The fourth car can have numbers 4 - 1111.

Total number of ways cars can be arranged = 1111!

The second part I am still trying to figure it out.
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 [#permalink] New post 14 Jun 2006, 01:44
is it 1 / ( 1108*1109*1110*1111) ?


Hi, buddy, thank you for remembering me :) This is a tricky one, i would say, especially, probability is not my strong point ;) . Anyway, i tried to solve and got that result. In fact, I'm sunk in tons of tasks right now, considering my up-coming G-day and 8 final papers early next month :cry:

Last edited by laxieqv on 14 Jun 2006, 03:27, edited 2 times in total.
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 [#permalink] New post 14 Jun 2006, 02:36
Am assuming increasing order to mean consecutive

Total number of ways for first four cars = 1111P4 = 1111*1110*1109*1108

Number of ways in which four cosecutive numbers can be chosen from 1-1111 = number of ways the first number can be chosen = 1108

P = 1108/(1111*1110*1109*1108)
= 1/1111*1110*1109

Am pretty sure the assumption stated upfront is invalid. Waiting for the OA!
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 [#permalink] New post 14 Jun 2006, 04:48
A shot in the dark. Is it 1/24?

The favorable event (that is the cars leaving the lot in increasing order) can be viewed as choosing 4 numbers from 1111 numbers in a certain order.
Hence favorable arrangements = 1111C4
All arrangements = 1111P4

Reqd probability = 1111C4/1111P4= 1/4! = 1/24
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 [#permalink] New post 14 Jun 2006, 05:24
giddi77 wrote:
A shot in the dark. Is it 1/24?

The favorable event (that is the cars leaving the lot in increasing order) can be viewed as choosing 4 numbers from 1111 numbers in a certain order.
Hence favorable arrangements = 1111C4
All arrangements = 1111P4

Reqd probability = 1111C4/1111P4= 1/4! = 1/24


I also think like that:

The problem can be drawn to getting the probability of a set of 4.
Because, for whichever the set of 4 is, there must be one and only one increasing order because the 4 numbers are different!
The probability for each increasing order is : 1/4! ( because such order is only one among 4! ways of arranging the set of 4)

But then i wrongly divided this by C(1111,4) . This is wrong , I did feel so. Now that i saw Giddi buddy has same direction; I can eliminate the ambiguousity :)

Continue:
There're C(1111,4) sets of 4 ---> the needed probability is:

1/4!* C(1111,4) / C(1111,4)
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 [#permalink] New post 14 Jun 2006, 06:59
Seems like a dependant probability question:

1/1111*1/1110*1/1109*1/1108 = 1/ some huge number 8-)
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 [#permalink] New post 14 Jun 2006, 07:08
No matter what group of four cars you choose, there is 1/4! chance that they are in increasing order. So the number 1111 really doesn't matter, isn't it?
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 [#permalink] New post 14 Jun 2006, 07:36
Total number of arrangements = 1111C4 * 4!
Only 1 combination from every set is in an increasing order.
Total number of increasing sets = 1111C4
Probability = 1/4!
------------------------------------------
Here's how I tested:

4 numbers: 1,2,3,4
# of ways to pick 3 numbers from the 4 = 4C3 = 4 --> (1,2,3) (1,3,4) (2,3,4) (1,2,4)

# of ways to arrange each set = 3! = 6
(1,2,3) = (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)

Total number of ways to arrange = 24 = 6 * 4 = 3! * 4C3

Note: each set only has 1 combination in increasing order (1,2,3) (1,3,4) (2,3,4) (1,2,4) --> That's your 4C3

So P = 4C3/(4C3 * 6)

I just substituted 4C3 with 111C3 and 3! with 4!
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 [#permalink] New post 14 Jun 2006, 07:48
HongHu wrote:
No matter what group of four cars you choose, there is 1/4! chance that they are in increasing order. So the number 1111 really doesn't matter, isn't it?


As usual your explanation is perfect! :good I somehow tend to think only in terms of nCr, nPr, n!.. formulas :beat
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 [#permalink] New post 14 Jun 2006, 12:06
laxieqv wrote:
is it 1 / ( 1108*1109*1110*1111) ?

Hi, buddy, thank you for remembering me :) This is a tricky one, i would say, especially, probability is not my strong point ;) . Anyway, i tried to solve and got that result. In fact, I'm sunk in tons of tasks right now, considering my up-coming G-day and 8 final papers early next month :cry:


thats :cool, laxi and goodluck to you..

thanx everybody. giddi, honghu, and ywilfred, all, did the goal.

it is 1/4! = 1/24
  [#permalink] 14 Jun 2006, 12:06
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