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:
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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
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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
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.
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.
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