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Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
02 Jul 2012, 19:15

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

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
27 Dec 2012, 01:17

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}

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
13 Jan 2013, 23:20

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. _________________

Re: PS - Prob. of A 4-letter code [#permalink]
27 Jan 2013, 14:50

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 _________________

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. _________________

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
05 May 2013, 22:10

Expert's post

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). _________________

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
10 Dec 2013, 03:25

1

This post received KUDOS

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.

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
30 Dec 2013, 04:17

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

OA is incorrect. Answer should be E as mentioned above

Slot method

3 choices for first slot 2 choices for second 1 choice for third 3 choice for fourth since once we have all of the letters then we can choose any of them

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
30 Dec 2013, 04:27

Expert's post

jlgdr wrote:

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

OA is incorrect. Answer should be E as mentioned above

Slot method

3 choices for first slot 2 choices for second 1 choice for third 3 choice for fourth since once we have all of the letters then we can choose any of them

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
31 Jul 2014, 00:17

Bunuel wrote:

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

Answer: C.

Hope it helps.

Could you please prove that "Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!" or show me the link that explain the formulation n!/r!? _________________

......................................................................... +1 Kudos please, if you like my post

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
31 Jul 2014, 01:20

Expert's post

vad3tha wrote:

Bunuel wrote:

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

Answer: C.

Hope it helps.

Could you please prove that "Total number of permutations for a set of 'n' objects of which 'r' objects are identical is n!/r!" or show me the link that explain the formulation n!/r!?

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
23 Aug 2014, 09:58

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.

Hi Bunuel,

Maybe I didn't read the question stem correctly, but at first, I did 3^4 (because 3 letters can go in each slot) -- this yields 81 which is not a choice. Then I realized that we need to have ABC(unknown) and my method can yield all A's or B's etc, therefore it's wrong. Is that the correct analysis?

That being said, I cannot come to terms with your formula above. How are we choosing 2 out of 4? isn't that what 4!/2! implies? Additionally, we don't really know where the unknown letter will go, it can go in slot 1, 2, 3, or 4. Can you please help me understand how you came up with the formula above?

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
22 Dec 2014, 05:10

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

Just adding to everyone, especially for folks who use MGMAT books. Please do correct me if I am wrong, I am still a noob.

Lets say we have to find number of ways the word MGMAT has to be arranged For such a question, we use \(5!/2!\)

Rephrasing this question, lets say we are told there is a 5 Letter code made from the letters "M,G,A,T" in which one of the letters repeats once - we do not know which one.

Then we use the following calculation

Arrangement in case M is repeated OR Arrangement in case G is repeated OR Arrangement in case A is repeated OR Arrangement in case T is repeated

\(5!/2!\) + \(5!/2!\) +\(5!/2!\) + \(5!/2!\)

Similarly, in this question we have 3 Letters in which we do not know which we know atleast one of them repeats but we do not know which one So Arrangement in case A is repeated OR Arrangement in case B is repeated OR Arrangement in case C is repeated

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
27 Oct 2015, 19:44

Bunuel wrote:

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

Answer: C.

Hope it helps.

Can someone please explain why 4!/2! represent the number of ways to arrange 4 letters of which one is repeated twice?? Thank you! _________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Re: A 4-letter code word consists of letters A, B, and C. If the [#permalink]
27 Oct 2015, 20:53

Expert's post

happyface101 wrote:

Bunuel wrote:

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

Answer: C.

Hope it helps.

Can someone please explain why 4!/2! represent the number of ways to arrange 4 letters of which one is repeated twice?? Thank you!

THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

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