joshi.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
Almost agree
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.
