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

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:

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

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;

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
_________________

I do not suffer from insanity. I enjoy every minute of it.

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

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A 4 letter code without constraint i.e, with 4 unique letters can be formed in 4! ways. Constraint is only 3 letters are unique and one of them repeats. With constraint the answer is 4! /2! =12 ways There are three such cases i.e., A ,B or C may repeat. So the total number of ways = 12*3=36
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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.

If you use the slot method, you can clearly see that the possible combinations will be

_ _ _ A or _ _ _ B or _ _ _ C

Thus no matter what set you choose, you will invariably end up with 2 same letters in any given combination of 4 letters.

Thus permutations of XYZX = 4!/2! and as you have 3 options to select A or B or C for the repetitive letter, the total number of arrangements possible = 4!/2!*3 = 36.

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