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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
 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. A-ABC can be arranged in 4!/2!=12 ways; B-ABC can be arranged in 4!/2!=12 ways; C-ABC 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 A-ABC, B-ABC. Why aren't we accounting for ABC-A , ABC-B . 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.
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
02 Jul 2012, 20:15
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
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
27 Dec 2012, 02:17
GHIBI wrote:
A 4-letter 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 4-letter 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
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
14 Jan 2013, 00:20
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.
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Re: PS - Prob. of A 4-letter code [#permalink]
27 Jan 2013, 15:50
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
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Re: PS - Prob. of A 4-letter code [#permalink]
27 Jan 2013, 21:20
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.
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
05 May 2013, 12:34
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 4-letter 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
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
05 May 2013, 23:10
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 4-letter 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 4-letter 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).
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
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. NEW!!!
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. NEW!!!
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Re: A 4-letter code word consists of letters A, B, and C. If the
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