Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 07 Dec 2006
Posts: 166

A 4letter code word consists of letters A, B, and C. If the [#permalink]
Show Tags
26 Jan 2008, 00:19
Question Stats:
31% (01:03) correct 69% (01:05) wrong based on 1593 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 47071

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
15 Sep 2010, 16:52
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.
_________________
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 Jan 2008
Posts: 45

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 08:23
It has to be 36 .. We know that the code already contains the the letters A,B and C. Now the 4th letter can be choosen in 3 ways (A,B,C). Once we have the set of 4 letters we can arrange them in 4!/2! ways. [Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!. if there are multiple groups of identical objects like say r1 objects which are red and r2 objects which are green .. then the total permutations would be n!/r1!r2! ...] Now the total ways of arranging would be 4!/2!*3 = 4*3*3 = 36 walker wrote: if you think N=36, try to add a new 3letters code to the 18set:
ABCA,ABCB,ABCC ACBA,ACBB,ACBC BACA,BACB,BACC BCAA,BCAB,BCAC CABA,CABB,CABC CBAA,CBAB,CBAC How about AACB ? ANSWER:C




CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 01:11
E\(N=C^3_1*C^2_1*C^1_1*C^3_1=3*2*1*3=18\) or \(N=3^4C^3_1C^3_1*C^2_1*C^4_1C^3_1*C^4_2*C^2_1*C^2_2=813*2*43*6*2*1=8132436=18\) or \(N=P^3_3*C^3_1=3*2*3=18\)
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Manager
Joined: 02 Jan 2008
Posts: 155

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 02:59
E: 18
3 letters A, B and C can be arranged in 3! ways = 6 ways 4th letter can be chosen in 3 ways
Total number of ways = 6*3=18



Manager
Joined: 01 Sep 2007
Posts: 95
Location: Astana

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 03:20
srp wrote: E: 18
3 letters A, B and C can be arranged in 3! ways = 6 ways 4th letter can be chosen in 3 ways
Total number of ways = 6*3=18 but then you assume that 4th letter can stand only in the end, don't you?



Manager
Joined: 01 Sep 2007
Posts: 95
Location: Astana

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 03:30
no, its not A. will be back in few moments
THE ANSWER SHOULD BE 36 WHICH IS HALF OF 72. CODE IS ESSENTIALLY A PERMUTATION, A1A2BC SHOULD BE THE SAME AS A2A1BC
3!*3*4/2= 36



Manager
Joined: 02 Jan 2008
Posts: 155

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 03:45
CaspAreaGuy wrote: srp wrote: E: 18
3 letters A, B and C can be arranged in 3! ways = 6 ways 4th letter can be chosen in 3 ways
Total number of ways = 6*3=18 but then you assume that 4th letter can stand only in the end, don't you? Right! so we then get ABC[R] where R stands for repeat letter of type A,B,C Number of ways of arranging ABC[R] = 4!/2! and since we have to do this 3 times for each A,B and C, Total ways = 4!/2! * 3 = 36



CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 05:52
if you think N=36, try to add a new 3letters code to the 18set: ABCA,ABCB,ABCC ACBA,ACBB,ACBC BACA,BACB,BACC BCAA,BCAB,BCAC CABA,CABB,CABC CBAA,CBAB,CBAC
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Manager
Joined: 01 Jan 2008
Posts: 220
Schools: Booth, Stern, Haas

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 06:07
Hi, CaspAreaGuy, nice to see you here CaspAreaGuy wrote: srp wrote: E: 18
3 letters A, B and C can be arranged in 3! ways = 6 ways 4th letter can be chosen in 3 ways
Total number of ways = 6*3=18 but then you assume that 4th letter can stand only in the end, don't you? I think that position of 4th letter doesn't matter 3* 2 * 1* 3=18 3* 3 * 1* 2 =18 3* 3* 2 * 1 =18



CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 10:13
You are right! +1 for Q and +1 for AACB live and learn....
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Jan 2008, 11:15
I corrected my formula: C\(N=\frac{C^3_3*C^3_1*P^4_4}{P^2_2}=\frac{1*3*4*3*2}{2}=36\) or \(N=3^4C^3_1C^3_1*C^2_1*C^4_1\frac12 *C^3_1*C^2_1*C^4_2=813*2*4\frac12 *3*6*2*1=8132418=36\)
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



SVP
Joined: 07 Nov 2007
Posts: 1732
Location: New York

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
26 Aug 2008, 08:56
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 ABCA + ABCB + ABCC = 4!/2! +4!/2!+4!/2! = 36
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Manager
Joined: 27 May 2009
Posts: 184

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
05 Sep 2009, 14:05
36 for me too. The group of 4 letters would be ABCX where X is A/B/C just find out the permutation for 1 specific case say ABCA 4 things can be permuted in 4! ways and since 2 things are same(here 2 A's) divide by 2! therefore... 4!/2! Since there are three such groups based upon value of X , multiply by 3. Ans: 3*(4!/2!) = 36
_________________
I do not suffer from insanity. I enjoy every minute of it.



Intern
Joined: 07 Aug 2009
Posts: 46

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
10 Sep 2009, 21:48
(4!/2) + (4!/2) + (4!/2 ) = 3*(4!/2 ) = 36 .
Ans: C
Explanation:
Assuming A is the repeated letter, we get the 1st 4!/2, OR if B is the repeated letter, we get the 2nd 4!/2 OR if C is the repeated letter, we get the 3rd 4!/2
that gives (4!/2) * 3 = 36



Manager
Joined: 11 Sep 2009
Posts: 129

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
11 Sep 2009, 16:34
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



Intern
Joined: 18 Aug 2010
Posts: 9

Re: PS  Prob. of A 4letter code [#permalink]
Show Tags
15 Sep 2010, 16:15
here is my approach: 3*2*1*3 = 18 but there is 4!/2! of arranging them =>36 ways in the end
_________________
D Day is April 23rd, 2010 Be humble, be focused, and be calm!



Manager
Joined: 26 Dec 2011
Posts: 99

Re: A 4letter code word consists of letters A, B, and C. If the [#permalink]
Show Tags
21 Jun 2012, 02:51
Here is how I tried to solve the question:
Consider XXXX = Assume the first three position is taken as for letters ABC those can be filled in 3! ways and then last letter can be filled in 3 ways... thus a total 18 ways.
Next, assume, that it XABC = 18 ways; next. CXAB = 18 ways; next BCXA = 18 ways... thus a total of 18*4 = 72ways!



Math Expert
Joined: 02 Sep 2009
Posts: 47071

Re: A 4letter code word consists of letters A, B, and C. If the [#permalink]
Show Tags
21 Jun 2012, 02:58



Math Expert
Joined: 02 Sep 2009
Posts: 47071

Re: A 4letter code word consists of letters A, B, and C. If the [#permalink]
Show Tags
22 Jun 2012, 00:39
arvind410 wrote: Bunuel wrote: pavanpuneet wrote: Here is how I tried to solve the question:
Consider XXXX = Assume the first three position is taken as for letters ABC those can be filled in 3! ways and then last letter can be filled in 3 ways... thus a total 18 ways.
Next, assume, that it XABC = 18 ways; next. CXAB = 18 ways; next BCXA = 18 ways... thus a total of 18*4 = 72ways! Note that the correct answer to this question is 36, not 72. AABC can be arranged in 4!/2!=12 ways; BABC can be arranged in 4!/2!=12 ways; CABC can be arranged in 4!/2!=12 ways; Total: 12+12+12=36. Answer: C. A quick question. In the above case , we assume that the extra letter is always in the first position , i.e AABC, BABC. Why aren't we accounting for ABCA , ABCB . And since it is a code, shouldn't arrangement matter, as in ABC is different from BAC ? Do correct me if I am wrong . We are not assuming that at all. The solution above says: AABC can be arranged in 4!/2!=12 ways. So, for the case when we have 2 A's in a code, 12 different arrangements are possible: AABC; ABAC; BAAC; ... The same for other cases. 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




Re: A 4letter code word consists of letters A, B, and C. If the
[#permalink]
22 Jun 2012, 00:39



Go to page
1 2 3
Next
[ 52 posts ]



