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# A 4-letter code word consists of letters A, B, and C. If the

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A 4-letter code word consists of letters A, B, and C. If the [#permalink]  25 Jan 2008, 23:19
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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: PS - Prob. of A 4-letter code [#permalink]  26 Aug 2008, 07:56
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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

ABCA + ABCB + ABCC

= 4!/2! +4!/2!+4!/2!
= 36
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Re: PS - Prob. of A 4-letter code [#permalink]  15 Sep 2010, 15:52
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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 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

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 $$3*\frac{4!}{2!}=36$$.

Hope it helps.
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 07:23
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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 3-letters code to the 18-set:

ABCA,ABCB,ABCC
ACBA,ACBB,ACBC
BACA,BACB,BACC
BCAA,BCAB,BCAC
CABA,CABB,CABC
CBAA,CBAB,CBAC

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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 01:59
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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
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Re: PS - Prob. of A 4-letter code [#permalink]  10 Sep 2009, 20:48
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(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
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]  21 Jun 2012, 01:58
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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.

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Re: PS - Prob. of A 4-letter code [#permalink]  05 Sep 2009, 13:05
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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
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 02:30
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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
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 10:15
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Expert's post
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

$$N=3^4-C^3_1-C^3_1*C^2_1*C^4_1-\frac12 *C^3_1*C^2_1*C^4_2=81-3*2*4-\frac12 *3*6*2*1=81-3-24-18=36$$
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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]  10 Dec 2013, 03:25
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Expert's post
Christy111 wrote:
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

First of all: 3!/2!=3, thus you'd have 3 + 3 + 3 = 9, not 12 + 12 + 12 = 36.

The number of ways to arrange AABC is 4!/2!=12. There is no need to consider AA as one unit, because we do not need AA to be together in each arrangement of AABC.

For complete solution check here: a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html#p782466

Hope it helps.
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A 4-letter code word consists of letters A, B, and C. If the [#permalink]  02 Nov 2015, 05:34
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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
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 00:11
Expert's post
E

$$N=C^3_1*C^2_1*C^1_1*C^3_1=3*2*1*3=18$$

or

$$N=3^4-C^3_1-C^3_1*C^2_1*C^4_1-C^3_1*C^4_2*C^2_1*C^2_2=81-3*2*4-3*6*2*1=81-3-24-36=18$$

or

$$N=P^3_3*C^3_1=3*2*3=18$$
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 02:20
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?
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 02:45
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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
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 04:52
Expert's post
if you think N=36, try to add a new 3-letters code to the 18-set:

ABCA,ABCB,ABCC
ACBA,ACBB,ACBC
BACA,BACB,BACC
BCAA,BCAB,BCAC
CABA,CABB,CABC
CBAA,CBAB,CBAC
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 05:07
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
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Re: PS - Prob. of A 4-letter code [#permalink]  26 Jan 2008, 09:13
Expert's post
You are right! +1 for Q and +1 for AACB

live and learn....
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Re: PS - Prob. of A 4-letter code [#permalink]  11 Sep 2009, 15:34
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
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Re: PS - Prob. of A 4-letter code [#permalink]  15 Sep 2010, 15:15
here is my approach:
3*2*1*3 = 18 but there is 4!/2! of arranging them =>36 ways in the end
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Re: PS - Prob. of A 4-letter code   [#permalink] 15 Sep 2010, 15:15

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