Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: [tex]3 P_{3}=(3)(2)(1)=6 [/tex]. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be [tex] \frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3 [/tex]. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.
