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: Combinatorics - Perms [#permalink]
17 Feb 2010, 02:21

bangalorian2000 wrote:

johnnymac wrote:

The answer is 13, not 35.

Question does not say that we cannot select duplicate letters. So to select 3 charaters from 7 characters hence 7C3.

This isn't correct. You need to take into account that some of the elements are duplicate. With your logic.. if u have the letters as {a,a,a,a,a,a,a}... then would no of comb possible for 3 letter word be 7c3? I don't think so.

Your answer would only be correct if all the 7 letters were different. If anyone of them repeated, the combinations would become less in number. Please check! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: How many combinations of three letters taken from letters [#permalink]
10 Aug 2012, 16:21

I always learn by modifying problems a little bit and see what answers I can come up with to test my understanding so...

If I assume that there's an extra a, an extra b, and an extra c for example. i.e. We have (a,a,a,b,b,b,c,c,c,d). Will that change our computation of 4C3+3C1*3C1? I myself don't think so but I'm waiting for your comments.

If I assume that's there's an an extra couple of D's and E's and 1 F i.e. (a,a,b,b,c,c,d,d,d,e,e,f). In this case the answer, as I guess, is 6C3+5C3*4C1=60. Am I correct?

Re: How many combinations of three letters taken from letters [#permalink]
23 Sep 2012, 07:22

Just got confused.

approaching by filling in slots.

we have three slots for seven letters.

now the first slot can be filled by any of the 7 letters, then 6, then 5.

so if there were 7 different letters, we would have had 7*6*5 and remove the repetitive combinations by dividing with 3! that makes 35 combinations.

For this approach, I got stuck here not knowing how to delete the repetitive combinations due to double letters. Help please.

P.S: I got the answer through other approach by adding unique and double letter combinations, I just want to understand why I got stuck above, rather how to continue from above.

Re: How many combinations of three letters taken from letters [#permalink]
25 Jan 2013, 11:36

Hey could someone please explain how they got 3C1 3C1? There are only two elements chosen in that situation? There doesn't seem to be a good explanation, or an OA for this problem _________________

Re: How many combinations of three letters taken from letters [#permalink]
26 Jan 2013, 04:09

6

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

bmwhype2 wrote:

How many combinations of three letters taken from letters (a, a, b, b, c, c, d) are possible?

A. 12 B. 13 C. 35 D. 36 E. 56

This kind of question has little chances appearing on the actual test.

Anyway, we have 7 letters {a, a, b, b, c, c, d}. There are 2 ways to select 3 letters out of this set:

CASE #1: all letters are distinct:

Since there are 4 distinct letters a, b, c and d, then the # of ways to select 3 out of 4 is 4C3=4.

CASE #2: 2 letters are the same and the third is different:

There are 3 letters from the set which can provide us with two letters: a, b, and c. 3C1=3 gives the # of ways to select which letter out of these 3 will provide us with 2 letters. For, example double letters can be aa, bb, or cc.

Next, we are left with 3 letters to choose the third letter. For example, if we choose aa, then b, c, and d are left to choose from for the third letter, thus the # of ways to do that is 3C1=3.

Total # of ways for this case is therefore 3C1*3C1=9.

Re: How many combinations of three letters taken from letters [#permalink]
26 Jul 2014, 21:54

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. _________________

Re: How many combinations of three letters taken from letters [#permalink]
24 Sep 2014, 06:57

bmwhype2 wrote:

How many combinations of three letters taken from letters (a, a, b, b, c, c, d) are possible?

A. 12 B. 13 C. 35 D. 36 E. 56

Sol:

consider any one pair with d one pair from 3 pairs = 3c1 *1 (AAD or BBD or CCD) But they can interchange their position(AAD,ADA,DAA) hence 3c1*1*3!/2! = 9

also we can consider all different digits

so we have 4 distinct objects and we have to choose 3 i.e. 4c3 = 4

Re: How many combinations of three letters taken from letters [#permalink]
06 Oct 2014, 04:44

johnnymac wrote:

jeeteshsingh - you use it to account for the following:

aa w/ b,c,d (aac, aab, aad) - 3 of these bb w/ a,c,d -3 of these cc w/ a,b,d -3 of these

3C1 is choose 3 letters w/ two of the same letter

It makes more sense to me as 3*3C1 = 9 total (rather than 3C1*3C1)

Indeed writing it as : 2C2*3C1*3=9 will make the most sense and clear it to almost anybody. ( 2C2 refers to the choosing of same 2 letters and then 3C1 refers to chossing any one letter from among the 3 diff types of letters left and then finally *3 gives the total such possible discrete cases.)

Hope this helps!! _________________

The Mind is everything . What you think you become. - Lord Buddha

Consider giving KUDOS if you appreciate my post !!

gmatclubot

Re: How many combinations of three letters taken from letters
[#permalink]
06 Oct 2014, 04:44

I´ve done an interview at Accepted.com quite a while ago and if any of you are interested, here is the link . I´m through my preparation of my second...

It has been a good week so far. After the disappointment with my GMAT score, I have started to study again, re-schedule the new test date and talked with...

It’s here. Internship season. The key is on searching and applying for the jobs that you feel confident working on, not doing something out of pressure. Rotman has...