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 500,000 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:

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 \(3*\frac{4!}{2!}=36\).

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:

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

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;

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.

A 4-letter code word consists of letters A, B, and C. If the [#permalink]

Show Tags

02 Nov 2015, 05:34

2

This post received KUDOS

A 4 letter code without constraint i.e, with 4 unique letters can be formed in 4! ways. Constraint is only 3 letters are unique and one of them repeats. With constraint the answer is 4! /2! =12 ways There are three such cases i.e., A ,B or C may repeat. So the total number of ways = 12*3=36
_________________

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.

If you use the slot method, you can clearly see that the possible combinations will be

_ _ _ A or _ _ _ B or _ _ _ C

Thus no matter what set you choose, you will invariably end up with 2 same letters in any given combination of 4 letters.

Thus permutations of XYZX = 4!/2! and as you have 3 options to select A or B or C for the repetitive letter, the total number of arrangements possible = 4!/2!*3 = 36.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...