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

You are right! +1 for Q and +1 for AACB

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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!?

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

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

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 ?

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?­