Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 28 Jul 2011
Posts: 404
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:
34% (01:39) correct 66% (01:27) wrong based on 251 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: 47898

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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




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: 129

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: 47898

Re: Find the number of ways in which 4 letters may be selected
[#permalink]
Show Tags
01 Jul 2013, 00:59



Manager
Joined: 14 Dec 2012
Posts: 69
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: 47898

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: \(C^4_8=70\)
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



SVP
Joined: 06 Sep 2013
Posts: 1852
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: 47898

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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



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: 7693

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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




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






