Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE. Aug 28 08:00 AM PDT  09:00 AM PDT Join a FREE live webinar with examPAL and Admissionado and learn how to master GMAT Critical Reasoning questions and the 6pointed star of MBA application essay glory. Save your spot today! Aug 30 08:00 PM PDT  11:00 PM PDT We'll be posting questions in DS/PS/SC/CR in competition mode. Detailed and quickest solution will get kudos. Will be collecting new links to all questions in this topic. Here you can also check links to fresh questions posted.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 28 Jul 2011
Posts: 323
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)

Find the number of ways in which 4 letters may be selected
[#permalink]
Show Tags
11 Jan 2012, 19:55
Question Stats:
35% (02:37) correct 65% (02:10) wrong based on 298 sessions
HideShow timer Statistics
Find the number of ways in which 4 letters may be selected from the word "Examination"? A. 66 B. 70 C. 136 D. 330 E. 4264 I was getting 142 but it is wrong,can anyone please help with this....??????
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
+1 Kudos If found helpful..




Math Expert
Joined: 02 Sep 2009
Posts: 57244

Re: Combinations
[#permalink]
Show Tags
26 Jan 2012, 11:02
raanan wrote: I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66 B. 70 C. 136 D. 330 E. 4264 We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X}; Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}. So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X} As pointed out, there are 3 different cases possible. {abcd}  all 4 letters are distinct: \(C^4_8=70\); {aabc}  two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left); {aabb}  two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs); Total=70+63+3=136. Answer: C. So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction. Hope it's clear.
_________________




Manager
Joined: 13 Dec 2011
Posts: 50
Location: United States (FL)
Concentration: Strategy, General Management
Schools: NYU Stern  Class of 2015
GMAT 1: 700 Q49 V37 GMAT 2: 740 Q48 V42
WE: Information Technology (Other)

Re: Combinations
[#permalink]
Show Tags
11 Jan 2012, 22:32
EXAMINATION has 11 letters, and in which 'A', 'I' and 'N', all occur twice. 11C4, would have been fine if all letters were distinct.
So, we have E, X, M, T, O, (AA), (II), (NN). 8 distinct letters.
1. 4 letters selected, which are all distinct: 8C4 = 70 2. 2 letters alike, and 2 distinct (eg: AAEX) = 3C1 x 7C2 = 63 3. 2 letters alike, and 2 letters alike (eg: AAII) = 3C2 = 3
So answer is, 70 + 63 + 3 = 136.
The tricky part is getting the second one correct, ( 2 letters alike, and 2 distinct)
Cheers...




Intern
Joined: 26 Jul 2011
Posts: 4

Re: Combinations
[#permalink]
Show Tags
11 Jan 2012, 21:13
Hi there, I got 330 and don't understand why this would be wrong. "examination" has 11 letters out of which we are selecting 4 letters. 11C4=11!/4!7!=330. What's the source of this question?



Manager
Status: Duke!
Joined: 20 Jan 2012
Posts: 127

Re: Combinations
[#permalink]
Show Tags
26 Jan 2012, 08:41
I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25



Math Expert
Joined: 02 Sep 2009
Posts: 57244

Re: Find the number of ways in which 4 letters may be selected
[#permalink]
Show Tags
01 Jul 2013, 00:59
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
_________________



Manager
Joined: 14 Dec 2012
Posts: 66
Location: United States

Re: Combinations
[#permalink]
Show Tags
24 Jul 2013, 14:01
Bunuel wrote: raanan wrote: I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66 B. 70 C. 136 D. 330 E. 4264 We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X}; Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}. So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X} As pointed out, there are 3 different cases possible. {abcd}  all 4 letters are distinct: \(C^4_8=70\); {aabc}  two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left); {aabb}  two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs); Total=70+63+3=136. Answer: C. So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction. Hope it's clear. Hi Bunuel, If the ques states that no repititions are possible, we will divide by factorial of no of time the letters are repeated to get the answer?Am i correct?Kindly clarify



Math Expert
Joined: 02 Sep 2009
Posts: 57244

Re: Combinations
[#permalink]
Show Tags
25 Jul 2013, 02:29
up4gmat wrote: Bunuel wrote: raanan wrote: I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66 B. 70 C. 136 D. 330 E. 4264 We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X}; Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}. So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X} As pointed out, there are 3 different cases possible. {abcd}  all 4 letters are distinct: \(C^4_8=70\); {aabc}  two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left); {aabb}  two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs); Total=70+63+3=136. Answer: C. So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction. Hope it's clear. Hi Bunuel, If the ques states that no repititions are possible, we will divide by factorial of no of time the letters are repeated to get the answer?Am i correct?Kindly clarify In this case we would be basically selecting 4 letters out of 8: {A, E, I, O, M, N, T, X}. Answer: \(C^4_8=70\)
_________________



SVP
Joined: 06 Sep 2013
Posts: 1629
Concentration: Finance

Re: Combinations
[#permalink]
Show Tags
30 May 2014, 05:50
Bunuel wrote: raanan wrote: I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66 B. 70 C. 136 D. 330 E. 4264 We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X}; Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}. So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X} As pointed out, there are 3 different cases possible. {abcd}  all 4 letters are distinct: \(C^4_8=70\); {aabc}  two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left); {aabb}  two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs); Total=70+63+3=136. Answer: C. So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction. Hope it's clear. When is it ok to use the anagram method for such problems? I did 11! / 5!2!2!2! , but obviously incorrect Any idea? Thanks! Cheers J



Math Expert
Joined: 02 Sep 2009
Posts: 57244

Re: Combinations
[#permalink]
Show Tags
30 May 2014, 07:20
jlgdr wrote: Bunuel wrote: raanan wrote: I would be thankful if someone could point out what is wrong in my solution: In total we have to select 4 out of 11, so its 11C4. each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations. counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 Find the number of ways in which 4 letters may be selected from the word "Examination"?A. 66 B. 70 C. 136 D. 330 E. 4264 We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X}; Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}. So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X} As pointed out, there are 3 different cases possible. {abcd}  all 4 letters are distinct: \(C^4_8=70\); {aabc}  two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left); {aabb}  two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs); Total=70+63+3=136. Answer: C. So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction. Hope it's clear. When is it ok to use the anagram method for such problems? I did 11! / 5!2!2!2! , but obviously incorrect Any idea? Thanks! Cheers J 11!/(2!2!2!) is the number of arrangements of the word "examination", which is not what the question is asking.
_________________



Intern
Joined: 04 Apr 2015
Posts: 16
Concentration: Human Resources, Healthcare
GMAT Date: 08062015
GPA: 3.83
WE: Editorial and Writing (Journalism and Publishing)

Re: Find the number of ways in which 4 letters may be selected
[#permalink]
Show Tags
03 Jul 2015, 23:29
I think the problem people are having with this question is in deciphering: This is a combination question where order of 'picking' does not matter. Then why do I've to group these similar letters. Well I had this problem too, this is how I solved it: 1. Understand what the question is asking: Is it asking about me 'picking' the letters? Or is it asking me about 'arranging' these letters differently? Ans: Its asking about 'picking'. So there you go, 1 question down 2. DONOT forget that the formula \(nCr\) is for n 'different' things  So here, you HAVE TO group the similar letter into sets. 3. Ok, now you've grouped them, what should be done now? Well Since, you've grouped them, you can have different cases of how u now 'pick' ! > Now, refer to what Bunuel said in his solution, and it'll make sense Hope this helps!!



NonHuman User
Joined: 09 Sep 2013
Posts: 12068

Re: Find the number of ways in which 4 letters may be selected
[#permalink]
Show Tags
11 Aug 2018, 13:38
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: Find the number of ways in which 4 letters may be selected
[#permalink]
11 Aug 2018, 13:38






