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Manager  Joined: 07 Dec 2006
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A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 33% (01:36) correct 67% (01:39) wrong based on 1758 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
Math Expert V
Joined: 02 Sep 2009
Posts: 65764
Re: PS - Prob. of A 4-letter code  [#permalink]

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29
32
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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Intern  Joined: 22 Jan 2008
Posts: 35
Re: PS - Prob. of A 4-letter code  [#permalink]

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16
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
CEO  B
Joined: 17 Nov 2007
Posts: 2913
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: PS - Prob. of A 4-letter code  [#permalink]

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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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Manager  Joined: 02 Jan 2008
Posts: 103
Re: PS - Prob. of A 4-letter code  [#permalink]

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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  Joined: 01 Sep 2007
Posts: 68
Location: Astana
Re: PS - Prob. of A 4-letter code  [#permalink]

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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: 68
Location: Astana
Re: PS - Prob. of A 4-letter code  [#permalink]

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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
Manager  Joined: 02 Jan 2008
Posts: 103
Re: PS - Prob. of A 4-letter code  [#permalink]

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2
1
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  B
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Re: PS - Prob. of A 4-letter code  [#permalink]

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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
Posts: 174
Schools: Booth, Stern, Haas
Re: PS - Prob. of A 4-letter code  [#permalink]

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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
CEO  B
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: PS - Prob. of A 4-letter code  [#permalink]

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You are right! +1 for Q and +1 for AACB live and learn.... _________________
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CEO  B
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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Re: PS - Prob. of A 4-letter code  [#permalink]

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2
2
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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VP  Joined: 07 Nov 2007
Posts: 1060
Location: New York
Re: PS - Prob. of A 4-letter code  [#permalink]

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11
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
Manager  Joined: 27 May 2009
Posts: 80
Re: PS - Prob. of A 4-letter code  [#permalink]

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6
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
Intern  Joined: 07 Aug 2009
Posts: 38
Re: PS - Prob. of A 4-letter code  [#permalink]

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5
(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: 108
Re: PS - Prob. of A 4-letter code  [#permalink]

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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: 8
Re: PS - Prob. of A 4-letter code  [#permalink]

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2
here is my approach:
3*2*1*3 = 18 but there is 4!/2! of arranging them =>36 ways in the end
Manager  Joined: 26 Dec 2011
Posts: 89
Re: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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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 V
Joined: 02 Sep 2009
Posts: 65764
Re: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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5
3
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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Math Expert V
Joined: 02 Sep 2009
Posts: 65764
Re: A 4-letter code word consists of letters A, B, and C. If the  [#permalink]

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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.
_________________ Re: A 4-letter code word consists of letters A, B, and C. If the   [#permalink] 21 Jun 2012, 23:39

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