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

We are given three letters, A, B, and C, and we must create a four-letter code in which all three letters are used. So, one letter must be repeated. Thus, we have the following three options:

1) A-B-C-A (if A is repeated)

2) A-B-C-B (if B is repeated)

3) A-B-C-C (if C is repeated)

Let’s start with option 1:

We see that there are four total letters and two repeated As. Thus, that code can be selected in the following number of ways:

4!/2! = (4 x 3 x 2 x 1)/(2 x 1) = 4 x 3 = 12 ways

Since the second code, A-B-C-B, has two Bs rather than two As, we can create the second code in 12 ways. Likewise, since the third code, A-B-C-C, has two Cs rather than two As or two Bs, we can create the third code in 12 ways.

Thus, the code can be created in 12 + 12 + 12 = 36 ways.

Answer: C

Sorry but I am really struggling to understand this:

-ABC can be arranged in 4!/2!=12

Many thanks

There are 4 letters not 3. For example, it says A-ABC can be arranged in 4!/2!=12 ways. AABC, so 4-letter out of which two A's are identical can be arranged in 4!/2!=12 ways.

To visualize:



The trap to avoid is that you don't only have to multiply the 3P3 by 3 possibilities for the 4th letter (which would be 18) but also multiply by 2 for the position (either in front or behind).

I will explain in detail :-
Using Fundamental Principle of counting
Assuming Four blocks are there _ _ _ _ which needs to filled by A,B,C.
1st can be filled in 3 ways (A,B,C)
2nd can be filled in 2 ways(2 remaining as one from A,B,C has already been given a place)
3rd can be filled in 1 ways (2 remaining as 2 from A,B,C have already been given a place and repetition is allowed )
4 can be filled in again 3 ways ( as we have already satisfied the condition for previous 3 spaces now this space can again be filled in 3 ways)
total outcomes :- 3*2*2*3=36
Is my reasoning correct in this case ?

No, you may be getting the answer, but you have missed out on arrangements where both As or Bs or Cs could be together.
For example AABC or BBAC.

Let me take another example- and we have 2 digits and A and B and we have to make 3 blocks.
As per your method
First any 2, next the one remaining and finally any of the two so 2*1*2=4

But actual method.
3 digits in which two are same so 3!/2=3 but the digit which is used twice can be any of the two, A or B, so 3*2=6

Calculating arrangements
AAB, ABA, BAA, BBA, BAB, ABB ——— 6 ways

But is the wording of the question a bit inprecise? If a code consists of A, B, or C, does it mean that there are no other letters present? Also, saying that the code includes all the 3 letters does not exclude the possibility that other letters are present, or am I wrong?! Additionally, is it always the case that in codes we are dealing with distinct letters or numbers unless otherwise mentioned?­

The phrase 'A 4-letter code word consists of letters A, B, and C' implies that the code contains only these letters. Regarding your other question, a properly formulated GMAT problem would clarify whether the symbols in the code are unique or distinct.