Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

There are 1111 cars in the parking lot each with number 1 to [#permalink]
13 Jun 2006, 21:55

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

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.

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

Last edited by laxieqv on 14 Jun 2006, 03:27, edited 2 times in total.

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 _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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:

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? _________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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

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! I somehow tend to think only in terms of nCr, nPr, n!.. formulas _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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

thats , laxi and goodluck to you..

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