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

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

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25 Jan 2008, 23:19
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Question Stats:

32% (01:05) correct 68% (01:06) wrong based on 1778 sessions

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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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Joined: 02 Sep 2009
Posts: 51215
Re: PS - Prob. of A 4-letter code  [#permalink]

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15 Sep 2010, 15:52
20
25
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]

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26 Jan 2008, 07:23
14
11
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

##### General Discussion
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Re: PS - Prob. of A 4-letter code  [#permalink]

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26 Jan 2008, 00:11
1
3
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]

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26 Jan 2008, 01:59
7
3
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
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Posts: 94
Location: Astana
Re: PS - Prob. of A 4-letter code  [#permalink]

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26 Jan 2008, 02:20
1
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]

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26 Jan 2008, 02:30
1
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]

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26 Jan 2008, 02:45
2
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]

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26 Jan 2008, 04:52
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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Manager
Joined: 01 Jan 2008
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Re: PS - Prob. of A 4-letter code  [#permalink]

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26 Jan 2008, 05:07
1
1
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]

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26 Jan 2008, 09:13
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]

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26 Jan 2008, 10:15
2
1
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: PS - Prob. of A 4-letter code  [#permalink]

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26 Aug 2008, 07:56
10
7
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]

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05 Sep 2009, 13:05
4
1
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]

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10 Sep 2009, 20:48
4
(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: PS - Prob. of A 4-letter code  [#permalink]

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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]

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15 Sep 2010, 15:15
1
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: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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21 Jun 2012, 01:51
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!
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Re: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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21 Jun 2012, 01:58
4
2
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: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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21 Jun 2012, 23:39
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.

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 &nbs [#permalink] 21 Jun 2012, 23:39

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