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
A-72
B-48
C-36
D-24
E-18
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Perm Comb Q
Source: Beat The GMAT — Problem Solving |
I am not sure of the OA.
Hence put in the forum
I am getting 18 but I feel I am wrong
To fill first place 3 options
To fill 2nd place 3 options
To fill 3rd place 2 options
To fill 4th place 1 option
Total = 3*3*2*1 = 18
Hence put in the forum
I am getting 18 but I feel I am wrong
To fill first place 3 options
To fill 2nd place 3 options
To fill 3rd place 2 options
To fill 4th place 1 option
Total = 3*3*2*1 = 18
I think it is 72.
Here are the details:
It is a 4 letter code and the order in which the alphabets are placed is also important hence there are 4! ways in which it can be done so 24 different ways in which this 4 letter code can be arranged.
Now in this case we have only 3 alphabets A,B,C so the 4th alphabet will be a repeat of any of the 3 alphabets. This combination of 4 letters will differ according the last letter used. so now there will be 3 different combinations possible for this.
24 X 3= 72
hence the ans is 72.
Any suggestions/corrections please
Thanks
Komal Joshi
Here are the details:
It is a 4 letter code and the order in which the alphabets are placed is also important hence there are 4! ways in which it can be done so 24 different ways in which this 4 letter code can be arranged.
Now in this case we have only 3 alphabets A,B,C so the 4th alphabet will be a repeat of any of the 3 alphabets. This combination of 4 letters will differ according the last letter used. so now there will be 3 different combinations possible for this.
24 X 3= 72
hence the ans is 72.
Any suggestions/corrections please
Thanks
Komal Joshi
sorry ,misread the question before ..answer is 36 ...since code contains all 3 letters one letter is repeated...
if we consider a is repeated we have 12 combinations..
6 combinations in which a is the first letter,3 combinations in which b is the first letter and 3 combinations in which c is the first letter...
so 12*3 (each of a, b and c being the repeat letter) combinations = 36
if we consider a is repeated we have 12 combinations..
6 combinations in which a is the first letter,3 combinations in which b is the first letter and 3 combinations in which c is the first letter...
so 12*3 (each of a, b and c being the repeat letter) combinations = 36
Almost agreejoshi.komal wrote:I think it is 72.
Here are the details:
It is a 4 letter code and the order in which the alphabets are placed is also important hence there are 4! ways in which it can be done so 24 different ways in which this 4 letter code can be arranged.
Now in this case we have only 3 alphabets A,B,C so the 4th alphabet will be a repeat of any of the 3 alphabets. This combination of 4 letters will differ according the last letter used. so now there will be 3 different combinations possible for this.
24 X 3= 72
hence the ans is 72.
Any suggestions/corrections please
Thanks
Komal Joshi
A different way to explain it might be to think of the 4th letter as X. Now we have 4! ways to permute A, B, C, X, which gives us 24. However, we know that X can be either A, B or C, which gives us 24 x 3 or 72 possibilities.
But remember that X is A, B or C, not a distinct letter. Thus certain permutations will not be distinct. I believe this will reduce the number of permutations by half. In general when there is 1 nondistinct letter in a group of k letters repeated n times, the permutation formula becomes k!/n!. Thus the answer would be 4! / 2! * 3 = 36
You could also bypass this formula by thinking about it. For each code, a letter will be repeated. So in writing out all the possible codes, each one will be duplicated as the two nondistinct letters can be swapped in position.
The code needs to contain A,B,C
so ABC can be arranged in 3! = 6ways
the 4th letter can be selected out of 3(A,B,C) in 3 ways
the 4th letter can be arranged in 4 ways with ABC
so total permuations are 6*3*4 =72
so ans should be 72
so ABC can be arranged in 3! = 6ways
the 4th letter can be selected out of 3(A,B,C) in 3 ways
the 4th letter can be arranged in 4 ways with ABC
so total permuations are 6*3*4 =72
so ans should be 72
Regards
Samir
Samir
