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: PS - Prob. of A 4-letter code [#permalink]
15 Sep 2010, 15:52

8

This post received KUDOS

Expert's post

8

This post was BOOKMARKED

MBAwannabe10 wrote:

here is my approach: 3*2*1*3 = 18 but there is 4!/2! of arranging them =>36 ways in the end

I don't see how the above way is giving 36 as an answer.

A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible? A. 72 B. 48 C. 36 D. 24 E. 18

As code must include all the three letters then pattern of the code word is ABCX where X can be any letter out of A, B, and C. So we can have the code word consisting of letters: ABCA; ABCB; ABCC.

We can arrange letters in each of above 3 cases in \(\frac{4!}{2!}\) # of ways (as each case has 4 letters out of which one is repeated twice), so total # of code words is \(3*\frac{4!}{2!}=36\).

Re: PS - Prob. of A 4-letter code [#permalink]
26 Jan 2008, 07:23

7

This post received KUDOS

2

This post was BOOKMARKED

It has to be 36 ..

We know that the code already contains the the letters A,B and C. Now the 4th letter can be choosen in 3 ways (A,B,C).

Once we have the set of 4 letters we can arrange them in 4!/2! ways.

[Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!. if there are multiple groups of identical objects like say r1 objects which are red and r2 objects which are green .. then the total permutations would be n!/r1!r2! ...]

Now the total ways of arranging would be 4!/2!*3 = 4*3*3 = 36

walker wrote:

if you think N=36, try to add a new 3-letters code to the 18-set:

Re: PS - Prob. of A 4-letter code [#permalink]
10 Sep 2009, 20:48

3

This post received KUDOS

(4!/2) + (4!/2) + (4!/2 ) = 3*(4!/2 ) = 36 .

Ans: C

Explanation:

Assuming A is the repeated letter, we get the 1st 4!/2, OR if B is the repeated letter, we get the 2nd 4!/2 OR if C is the repeated letter, we get the 3rd 4!/2

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
21 Jun 2012, 01:58

3

This post received KUDOS

Expert's post

pavanpuneet wrote:

Here is how I tried to solve the question:

Consider XXXX = Assume the first three position is taken as for letters ABC those can be filled in 3! ways and then last letter can be filled in 3 ways... thus a total 18 ways.

Next, assume, that it XABC = 18 ways; next. CXAB = 18 ways; next BCXA = 18 ways... thus a total of 18*4 = 72ways!

Note that the correct answer to this question is 36, not 72.

A-ABC can be arranged in 4!/2!=12 ways; B-ABC can be arranged in 4!/2!=12 ways; C-ABC can be arranged in 4!/2!=12 ways;

Re: PS - Prob. of A 4-letter code [#permalink]
05 Sep 2009, 13:05

2

This post received KUDOS

36 for me too. The group of 4 letters would be ABCX where X is A/B/C just find out the permutation for 1 specific case say ABCA 4 things can be permuted in 4! ways and since 2 things are same(here 2 A's) divide by 2! therefore... 4!/2! Since there are three such groups based upon value of X , multiply by 3. Ans: 3*(4!/2!) = 36 _________________

I do not suffer from insanity. I enjoy every minute of it.

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
10 Dec 2013, 03:25

1

This post received KUDOS

Expert's post

Christy111 wrote:

My approach is the following: aabc - 3!2! (aa as one unit, which gives 3!, b and c interchangeable, which is 2!) bbac - 3!2! ccab - 3!2!

12+12+12=36 Is it correct? Thank you

First of all: 3!/2!=3, thus you'd have 3 + 3 + 3 = 9, not 12 + 12 + 12 = 36.

The number of ways to arrange AABC is 4!/2!=12. There is no need to consider AA as one unit, because we do not need AA to be together in each arrangement of AABC.

Re: PS - Prob. of A 4-letter code [#permalink]
11 Sep 2009, 15:34

I get C: 36 as well. This is how I approached the problem:

The 4 letters can be distinguished as follows: X - the letter which is duplicated. Y and Z - the two remaining letters, with Y always preceding Z in the code word.

As a result, a code word looks like XXYZ, or XYXZ, etc.

The number of possible combinations is as follows:

4C2 - choose 2 of the 4 character places to put the duplicate characters (X in this case) * 3! - 3 ways to choose X, 2 ways to choose Y, 1 way to choose Z.

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
21 Jun 2012, 01:51

Here is how I tried to solve the question:

Consider XXXX = Assume the first three position is taken as for letters ABC those can be filled in 3! ways and then last letter can be filled in 3 ways... thus a total 18 ways.

Next, assume, that it XABC = 18 ways; next. CXAB = 18 ways; next BCXA = 18 ways... thus a total of 18*4 = 72ways!

gmatclubot

Re: A 4-letter code word consists of letters A, B, and C. If the
[#permalink]
21 Jun 2012, 01:51

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...