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:

Re: Permuation Problem [#permalink]
26 Mar 2008, 16:51

3

This post received KUDOS

I figured out the solution however it is not a simple permutation question. This includes both permutation as well as combination.

There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice.

Now first we need to see how many ways we can make word with 4 letter between P and S. Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210

Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S. So total way = 210*2 = 420

The selected 4 letters can be rotated between P and S in = 4! ways

So total ways = 420 * 4!

Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter. Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways.

So total number of ways = 7! * 420 * 4!

Now since letter T was repeated twice, we should divide the above result by 2!.

So Total number of ways = 7! * 420 * 4! / 2! = 25401600

Re: Permuation Problem [#permalink]
24 Mar 2010, 18:51

1

This post received KUDOS

Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Re: Permuation Problem [#permalink]
25 Mar 2010, 02:27

crack700 wrote:

Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more? _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: Permuation Problem [#permalink]
25 Mar 2010, 04:34

2

This post received KUDOS

AtifS wrote:

crack700 wrote:

Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

. . . . . _ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

Re: Permuation Problem [#permalink]
25 Mar 2010, 08:32

sudgmat wrote:

AtifS wrote:

crack700 wrote:

Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is 14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

. . . . . _ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

Thanks

Ravi

Thanks! man. kudos _________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??

Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]
10 Nov 2013, 02:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

Back to the original questions:

In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?

There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice.

1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210; 2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2; 3. Permutation of the 4 letters between P and S = 4! =24; 4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040; 5. We should divide multiplication of the above 4 numbers by 2! as there is repeated T.

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

For the past couple of weeks I’ve been winding down my affairs in New York by working on consulting projects, trying every exotic sandwich there is and then intensely...