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Re: A 4letter code word consists of letters A, B, and C. If the
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04 Feb 2016, 06:34
If the question didnt involve the restriction "If the code includes all the three letters", would the answer be 3^4=81? I am not able to imagine how that should be the answer. Can some one help? thanks



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Re: A 4letter code word consists of letters A, B, and C. If the
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04 Feb 2016, 06:42
BrainLab wrote: can somebody solve it with a slot method ? 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. D is thus the correct answer. Hope this helps.



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Re: A 4letter code word consists of letters A, B, and C. If the
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04 Feb 2016, 06:42
Nina1987 wrote: If the question didnt involve the restriction "If the code includes all the three letters", would the answer be 3^4=81? I am not able to imagine how that should be the answer. Can some one help? thanks 3^4 is only possible if you are given that the 4 letter code can have repetitive letters such as AAAA or BBBB or CCCC etc.



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Re: A 4letter code word consists of letters A, B, and C. If the
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05 Jul 2016, 08:24
Can anyone help me with this one? This is my reasoning: Total combination of codes: 3*3*3*3=81 Cases of only twoletter codes: 2*2*2*22 (the 2 if for the cases that only 1 letter repeats 4 times)(could be any of the two letters). Cases of oneletter codes: 3 (AAA, BBB or CCC).
Answer: 81143= 64
Makes sense to me!



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Re: A 4letter code word consists of letters A, B, and C. If the
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09 Jul 2016, 00:06
A/c to the question,we have to make a 4 letter word using the letters A,B,C only and all letters have to be used.So,there are 3 possible cases:
Case 1: Using A twice,B once,C once
How many different codes can be formed using the letters A,A,B,C? = 4!/2!= 12
Explanation: N different objects can arrange in n different places in n! ways.So,these 4 letters arrange in 4 places in 4! ways if these all were different.Since, 2 letters are same ,the actual no. of arrangements would be 4!/2!
Case 2: Using B twice,A once,C once
Similar to Case 1,Total no. of arrangements would be 12
Case 3 : Using C twice,A once,B once
Similar to Case 1,Total no. of arrangements would be 12
So,Total arrangements would be 36
Hope that helps!



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A 4letter code word consists of letters A, B, and C. If the
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11 Jul 2016, 05:11
GHIBI wrote: A 4letter 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 Take one case: AABC With two A's together, there are 6 ways of forming a code. Separate A's by one letter. There are 4 ways of forming a code. Separate A's by 2 letters . There are 2 ways of forming a code. A total of 12 ways. Similarly for B and C repeating . Final total of 36.
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Re: A 4letter code word consists of letters A, B, and C. If the
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02 Aug 2016, 22:13
Well the question is a bit ambiguous as it is not clear whether the 4 letter word is to formed ONLY by the 3 letters or the 4th letter can be different also In the solution given below I have considered that the 4 letter word is formed only by the 3 letters.Hence in this case it obvious that each letter will be repeated twice for each combination therefore for 4!/(2!)*3=36 IS THE ANSWER



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Re: A 4letter code word consists of letters A, B, and C. If the
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10 May 2017, 10:33
GHIBI wrote: A 4letter 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 A bit logic is needed for this problem.there are three letters and we need to form 4 letters code. _ _ _ (A,B,or C).For every code we have to repeat one letter. So, the solution will be : 3*4/2! = 36 . hope it helps



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Re: A 4letter code word consists of letters A, B, and C. If the
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16 May 2017, 17:16
GHIBI wrote: A 4letter 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 We are given three letters, A, B, and C, and we must create a fourletter code in which all three letters are used. So, one letter must be repeated. Thus, we have the following three options: 1) ABCA (if A is repeated) 2) ABCB (if B is repeated) 3) ABCC (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, ABCB, has two Bs rather than two As, we can create the second code in 12 ways. Likewise, since the third code, ABCC, 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
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Re: A 4letter code word consists of letters A, B, and C. If the
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04 Jun 2017, 08:04
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. AABC can be arranged in 4!/2!=12 ways; BABC can be arranged in 4!/2!=12 ways; CABC can be arranged in 4!/2!=12 ways; Total: 12+12+12=36. Answer: C. Sorry but I am really struggling to understand this: ABC can be arranged in 4!/2!=12 Many thanks



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Re: A 4letter code word consists of letters A, B, and C. If the
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04 Jun 2017, 22:31
Pol DC wrote: 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. AABC can be arranged in 4!/2!=12 ways; BABC can be arranged in 4!/2!=12 ways; CABC can be arranged in 4!/2!=12 ways; Total: 12+12+12=36. 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 AABC can be arranged in 4!/2!=12 ways. AABC, so 4letter out of which two A's are identical can be arranged in 4!/2!=12 ways.
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A 4letter code word consists of letters A, B, and C. If the
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29 Dec 2017, 17:09
If a 4 letter code has all the 3 letters A B and C, that means that 1 letter is repeated. Let's say it is repeated in position 1 and 2, the number of possibilities is 3 * 1 * 2 * 1= 6, because once we have the first letter, we only have one possibility for the second one. Now we only need to know how many combinations of 2 positions we can have in 4 positions = C(4,2) =6 So the answer is 6 * 6 = 36 C



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Re: A 4letter code word consists of letters A, B, and C. If the
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24 Oct 2018, 04:45
It took me a little while to digest this one. Here is how i understood it
I broke the question into 2 parts 1. Select the letters AND 2. Arrange the letters
1. Select the letters we know that 3 of the 4 digits have to be A B C. They can be selected in 1 way (3C3). The 4th digit is going to be one of 3 (A B C) which can be selected in 3 ways (3C1). Hold onto this.
2. Arranging the letters Now we have 4 digits that need to be arranged. We also know that one of these digits is repeating itself twice. It will be 4!/2! = 12.
Now we just use AND to combine these two. 1*3*12 = 36.



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Re: A 4letter code word consists of letters A, B, and C. If the
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02 Nov 2018, 07:56
Idea : 3 Letters but 4 places to be arranged. So there has to be a repetition of one of the letters exactly once assuming the code words does not contain any other elements.
Approach : What is the output : Code word. so order matters.Also there are 4 positions. So we need to use combinations for each possibility of placement of three letters. So there are 4C3 ways of placing these three letters among these 4 places and in each case order matters , so 3! ways of arranging these three letters . Now the letter in the fourth position in each case has to be one of three letters so we have three choices.So total possible cases are: 4C3 * 3! * 3 = 72 ways. However we will have repetition of one letter in each cases so by symmetry property we have to slice it into 2! ways. So the number of distinct possible code words: 72 / 2 = 36 ways.




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