January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 Dec 2009
Posts: 66
Location: india

How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
14 Apr 2010, 04:33
Question Stats:
50% (01:38) correct 50% (01:01) wrong based on 45 sessions
HideShow timer Statistics
How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.




Math Expert
Joined: 02 Sep 2009
Posts: 52285

Re: tough p n c
[#permalink]
Show Tags
14 Apr 2010, 05:45
jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454.
_________________
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: 07 Apr 2010
Posts: 21

Re: tough p n c
[#permalink]
Show Tags
14 Apr 2010, 06:40
a very similar question: Find the no: of 4 letter words that can be formed from the string "AABBBBCC" ? Here we have 3 distinct letters(A,B,C) & 4 slots to fill. What logic do you use to solve this problem?
_________________
GMAT, here i come... impressed?...how bout encouraging me with a kudos
cheers just another idiot!



Math Expert
Joined: 02 Sep 2009
Posts: 52285

Re: tough p n c
[#permalink]
Show Tags
14 Apr 2010, 07:01



Intern
Joined: 07 Apr 2010
Posts: 21

Re: tough p n c
[#permalink]
Show Tags
14 Apr 2010, 10:41
thanks a ton, bunuel
_________________
GMAT, here i come... impressed?...how bout encouraging me with a kudos
cheers just another idiot!



Manager
Joined: 21 Mar 2010
Posts: 94

Re: tough p n c
[#permalink]
Show Tags
15 Apr 2010, 06:20
I'm usually not bad with anagram problems like this but the term "words" threw me off completely. For some reason I assumed the combination of letters had to combine to make sense, i.e. a "word".
MTHE  is hardly a word, so i started counting actual "words"... so obviously completely bombed the question!



Manager
Joined: 14 Nov 2011
Posts: 122
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: tough p n c
[#permalink]
Show Tags
25 May 2013, 17:50
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Hi Bunnel, Is this a GMAT worthy question?



Math Expert
Joined: 02 Sep 2009
Posts: 52285

Re: tough p n c
[#permalink]
Show Tags
26 May 2013, 03:13
cumulonimbus wrote: Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Hi Bunnel, Is this a GMAT worthy question? No, but this question is good to practice.
_________________
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: 22 Mar 2013
Posts: 11

Re: tough p n c
[#permalink]
Show Tags
02 Jul 2013, 10:45
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Bunuel, this is a damn hard question and I find myself not fully able to understand your logic. I am from a very weak background but I have poured through all of the MGMAT math books (excluding the advanced one) several times and still find myself unable to intutively figure out the steps to this problem. What extra review would you suggest so I can be able to at least follow your solutions to these answers?



Math Expert
Joined: 02 Sep 2009
Posts: 52285

Re: tough p n c
[#permalink]
Show Tags
02 Jul 2013, 11:21
tmipanthers wrote: Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Bunuel, this is a damn hard question and I find myself not fully able to understand your logic. I am from a very weak background but I have poured through all of the MGMAT math books (excluding the advanced one) several times and still find myself unable to intutively figure out the steps to this problem. What extra review would you suggest so I can be able to at least follow your solutions to these answers? This question is out of the scope of the GMAT, so I wouldn't worry about it too much. As for the recommendations. Best GMAT Books: bestgmatmathprepbooksreviewsrecommendations77291.htmlTheory on Combinations: mathcombinatorics87345.htmlDS questions on Combinations: search.php?search_id=tag&tag_id=31PS questions on Combinations: search.php?search_id=tag&tag_id=52Tough and tricky questions on Combinations: hardestareaquestionsprobabilityandcombinations101361.htmlHope it helps.
_________________
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: 27 Mar 2013
Posts: 42
Location: United States
Concentration: Strategy, Entrepreneurship
GPA: 3.25
WE: General Management (Energy and Utilities)

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
24 Mar 2014, 12:36
If this would have been a word with three of the same letter I'm assuming you would have more than 3 combinations?
Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 52285

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
25 Mar 2014, 01:43



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: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
10 Jul 2015, 10:44
I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements?



CEO
Joined: 20 Mar 2014
Posts: 2638
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
10 Jul 2015, 11:03
ddg wrote: I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements? You are half right. Permutations = Combinations * n! (where n is the number of 'elements'). In this question, you first need to select the letters out of the given one (combination implied as selection = combination!!) and only after you have selected the letters , you can look at the arrangements. You can not directly go to arrangements as you need to follow the 2 step process: 1. Choose 4 out of 11 letters 2. Arrangement of those selections of 4 letters to get all the possible arrangements. Your approach would have been correct, had the question ask us to arrange all of these 11 letters into words of 11 letters or if all the letters were different.



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: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
10 Jul 2015, 12:24
Thanks! this helped



Retired Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 497

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
28 Nov 2015, 11:14
Bunuel wrote: idiot wrote: a very similar question: Find the no: of 4 letter words that can be formed from the string "AABBBBCC" ?
Here we have 3 distinct letters(A,B,C) & 4 slots to fill. What logic do you use to solve this problem? Three patterns: 1. XXXX  only BBBB, so 1 2. XXYY  3C2(choosing which will take the places of X and Y from A, B and C)*4!/2!2!(arranging)=18 3. XXYZ  3C1(choosing which will take the place of X from A, B and C)*4!/2!(arranging)=36 4. XXXY  2C1(choosing which will take the place of Y from A and C, as X can be only B)*4!/3!(arranging)=8 1+18+36+8=63 In 3, through 3C1 we choose 1 group of letters from A, B, and C, which will take the place of XX, but what about the remaining 2 letters YZ? One more query, if through 3C1 we choose group B, then we have 4 letters in hand unlike A and C. Doesn't it require to consider?
_________________
Hasan Mahmud



Intern
Joined: 31 Aug 2016
Posts: 46

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
06 Dec 2017, 11:36
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. In (1) why you use order matters or doesn't matter to solve it? To saw two different ways to find the solution?



Intern
Joined: 13 Dec 2016
Posts: 1

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
Show Tags
29 Oct 2018, 17:02
Because there's no restriction, why can't I simply take 11*10*9*8=7920? What is wrong with this logic?




Re: How many words can be formed by taking 4 letters at a time &nbs
[#permalink]
29 Oct 2018, 17:02






