January 17, 2019 January 17, 2019 08:00 AM PST 09:00 AM PST Learn the winning strategy for a high GRE score — what do people who reach a high score do differently? We're going to share insights, tips and strategies from data we've collected from over 50,000 students who used examPAL. 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.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 20 Sep 2011
Posts: 23

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
02 Jul 2012, 19:15
What about this method? I used the slot method to answer this one. XXXX  First slot you have three options, second slot you have three options (nothing says you can't repeat letters), third slot you have two options and fourth slot you have two options to make sure that you include at least all of the letters. 3*3*2*2 = 36



Senior Manager
Joined: 13 Aug 2012
Posts: 426
Concentration: Marketing, Finance
GPA: 3.23

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
27 Dec 2012, 01:17
GHIBI wrote: A 4letter 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
So, the 4letter code will have A,B,C and a repeat letter from either A,B or C. Our possible selections could be: {A,A,B,C}, {B,B,A,C}, and {C,C,A,B}\(A,A,B,C > 4!/2! = 12\) \(B,B,A,C > 4!/2! = 12\) \(C,C,B,A > 4!/2! = 12\) Answer: 36
_________________
Impossible is nothing to God.



Manager
Joined: 12 Jan 2013
Posts: 56
Location: United States (NY)
GPA: 3.89

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
13 Jan 2013, 23:20
First pick which letter is doubled. There are 3 ways. Without loss of generality, A is doubled. Then pick a place for B. There are 4 ways to pick a place for B. Then pick a place for C. There are 3 ways to pick a place for C. Then place the two A's. 3*4*3=36. It is similar to AKProdigy87's solution, except that we don't even need the 4C2. No formula whatsoever, just multiplication.
_________________
Sergey Orshanskiy, Ph.D. I tutor in NYC: http://www.wyzant.com/Tutors/NY/NewYork/7948121/#ref=1RKFOZ



Manager
Joined: 07 Feb 2011
Posts: 89

Re: PS  Prob. of A 4letter code
[#permalink]
Show Tags
27 Jan 2013, 14:50
AKProdigy87 wrote: I get C: 36 as well. This is how I approached the problem:
The 4 letters can be distinguished as follows: X  the letter which is duplicated. Y and Z  the two remaining letters, with Y always preceding Z in the code word.
As a result, a code word looks like XXYZ, or XYXZ, etc.
The number of possible combinations is as follows:
4C2  choose 2 of the 4 character places to put the duplicate characters (X in this case) * 3!  3 ways to choose X, 2 ways to choose Y, 1 way to choose Z.
4C2 * 3! = 36 What I don't get about this approach is why we don't multiply by 2! to account for the different permutations of YZ, and hence have an answer of 72. I know it's an old problem, but would someone care to explain? Where is this permutation of 2! for YZ already accounted for in this problem? That's really unclear to me
_________________
We appreciate your kudos'



Manager
Joined: 12 Jan 2013
Posts: 56
Location: United States (NY)
GPA: 3.89

Re: PS  Prob. of A 4letter code
[#permalink]
Show Tags
27 Jan 2013, 20:20
manimgoindowndown wrote: I know it's an old problem, but would someone care to explain?
Where is this permutation of 2! for YZ already accounted for in this problem? That's really unclear to me AABC, ABAC, ABCA, BAAC, BACA, BCAA, AACB, ACAB, ACBA, CAAB, CABA, CBAA BBAC, BABC, BACB, ABBC, ABCB, ACBB, BBCA, BCBA, BCAB, CBBA, CBAB, CABB CCAB, CACB, CABC, ACCB, ACBC, ABCC, CCBA, CBCA, CBAC, BCCA, BCAC, BACC We agree that Y always preceeds Z. Then we have two ways to choose Y. For example, if B repeats twice, then we may have Y=A (e.g. BBAC) or Y=C (e.g. BBCA). Y and Z cannot be permuted.
_________________
Sergey Orshanskiy, Ph.D. I tutor in NYC: http://www.wyzant.com/Tutors/NY/NewYork/7948121/#ref=1RKFOZ



Intern
Joined: 20 Apr 2013
Posts: 20
Concentration: Finance, Finance
GMAT Date: 06032013
GPA: 3.3
WE: Accounting (Accounting)

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
05 May 2013, 11:34
Bunuel Please Clarify my doubt. If the code includes all the three letters, then the 4th letter can any letter from ABCDEFG.... XYZ. The question doesn't specify that the 4 letter code includes only A, B and C. GHIBI wrote: A 4letter 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



Math Expert
Joined: 02 Sep 2009
Posts: 52161

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
05 May 2013, 22:10
Rajkiranmareedu wrote: Bunuel Please Clarify my doubt. If the code includes all the three letters, then the 4th letter can any letter from ABCDEFG.... XYZ. The question doesn't specify that the 4 letter code includes only A, B and C. GHIBI wrote: A 4letter 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 I think that it is specified. We are told that a 4letter code consists of letters A, B, and C and that the code includes ALL the three letters A, B, and C (so the case of AAAA is not possible).
_________________
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: 19 Mar 2013
Posts: 22

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
10 Dec 2013, 03:15
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



Math Expert
Joined: 02 Sep 2009
Posts: 52161

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
10 Dec 2013, 03:25



Intern
Joined: 17 Jan 2014
Posts: 7

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
18 Jan 2014, 19:25
Another solution: P(3,3)C(3,1)P(2,1)=36
(C)



Manager
Joined: 22 Feb 2009
Posts: 168

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
31 Jul 2014, 00:17
Bunuel wrote: MBAwannabe10 wrote: 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 4letter 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 \(3*\frac{4!}{2!}=36\). Answer: C. Hope it helps. Could you please prove that "Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!" or show me the link that explain the formulation n!/r!?
_________________
......................................................................... +1 Kudos please, if you like my post



Math Expert
Joined: 02 Sep 2009
Posts: 52161

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
31 Jul 2014, 01:20
vad3tha wrote: Bunuel wrote: MBAwannabe10 wrote: 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 4letter 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 \(3*\frac{4!}{2!}=36\). Answer: C. Hope it helps. Could you please prove that "Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!" or show me the link that explain the formulation n!/r!? Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word. Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
_________________
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: 13 Dec 2013
Posts: 40
GPA: 2.71

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
22 Dec 2014, 05:10
GHIBI wrote: A 4letter 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 Just adding to everyone, especially for folks who use MGMAT books. Please do correct me if I am wrong, I am still a noob. Lets say we have to find number of ways the word MGMAT has to be arranged For such a question, we use \(5!/2!\) Rephrasing this question, lets say we are told there is a 5 Letter code made from the letters "M,G,A,T" in which one of the letters repeats once  we do not know which one. Then we use the following calculation Arrangement in case M is repeated OR Arrangement in case G is repeated OR Arrangement in case A is repeated OR Arrangement in case T is repeated \(5!/2!\) + \(5!/2!\) +\(5!/2!\) + \(5!/2!\) Similarly, in this question we have 3 Letters in which we do not know which we know atleast one of them repeats but we do not know which one So Arrangement in case A is repeated OR Arrangement in case B is repeated OR Arrangement in case C is repeated \(4!/2!\) + \(4!/2!\) +\(4!/2!\) =36



Intern
Joined: 05 Aug 2015
Posts: 42

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
27 Oct 2015, 19:44
Bunuel wrote: MBAwannabe10 wrote: 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 4letter 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 \(3*\frac{4!}{2!}=36\). Answer: C. Hope it helps. Can someone please explain why 4!/2! represent the number of ways to arrange 4 letters of which one is repeated twice?? Thank you!
_________________
Working towards 25 Kudos for the Gmatclub Exams  help meee I'm poooor



Math Expert
Joined: 02 Sep 2009
Posts: 52161

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
27 Oct 2015, 20:53
happyface101 wrote: Bunuel wrote: MBAwannabe10 wrote: 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 4letter 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 \(3*\frac{4!}{2!}=36\). Answer: C. Hope it helps. Can someone please explain why 4!/2! represent the number of ways to arrange 4 letters of which one is repeated twice?? Thank you! THEORY:Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is: \(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\). For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word. Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice. Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\). For more check the links below: Combinatorics Made Easy!Theory on CombinationsDS questions on CombinationsPS questions on CombinationsTough and tricky questions on Combinations
_________________
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



Director
Joined: 10 Mar 2013
Posts: 503
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
28 Oct 2015, 02:11
can somebody solve it with a slot method ?
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Intern
Joined: 28 Aug 2015
Posts: 19
Location: Norway
WE: Information Technology (Consulting)

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
31 Oct 2015, 14:03
BrainLab wrote: can somebody solve it with a slot method ? Hi, Four letter code, let's assume ABCX, X any letter "A or B or C". As you want to solve this with slot method, "ABC" can be arranged in 3!. Now come to ABCX : A can be arranged in any 4 way. Similarly B can arrange in any 3 way, C can arrange in 2 way and Last "X" which comprises of (A or B or C). Let suppose if you put "A" again at forth place this can done in 1 way as one slot is remaining to fill ABCA  > 4*3*2 *1 = 24.But as "A" is repeated twice, we need to remove this ambiguity. We divide this by 2, which is 24/2 = 12. Now, Total combination is 12*3 = 36, which is correct answer. Hope this will help you Thanks. If you like my post Kudos Please
_________________
Thanks Sumit kumar If you like my post Kudos Please



Director
Joined: 17 Dec 2012
Posts: 625
Location: India

A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
02 Nov 2015, 05:34
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
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Current Student
Joined: 02 Jun 2015
Posts: 81
Location: Brazil
Concentration: Entrepreneurship, General Management
GPA: 3.3

A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
02 Nov 2015, 05:37
First we need to think about the words that are going to repeat.
1st case: ABCA (you have two letters A) 4!/2! = 12 (2! is used because the letter A repeats 2 times)
2nd case: ABCB 4!/2! = 12
3rd case: ABCC 4!/2! = 12
12*3= 36



Intern
Joined: 08 Dec 2015
Posts: 23

Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
Show Tags
14 Dec 2015, 13:33
Honestly, this is a really tough topic. I calculated it this way:
ABCX + ABXC + AXBC + XABC then divide by 2 because of the repeat (i.e. ABCC and ABCC would show up in the first term and the second term?
3*2*1*3 + 3*2*3*1 + 3*3*2*1 + 3*3*2*1 = 72 > 72/2 = 36
Can you please tell me if this is a correct line of thinking. Or better yet, how to develop any type of intuition on these problems?




Re: A 4letter code word consists of letters A, B, and C. If the &nbs
[#permalink]
14 Dec 2015, 13:33



Go to page
Previous
1 2 3
Next
[ 54 posts ]



