Micheal arranges the letters in the word BANANA randomly. What is the probability that the As are all next to each other?

Question

Micheal arranges the letters in the word BANANA randomly. What is the probability that the As are all next to each other?

What I did : So since there are 6 letters there are 6! or 720 possibilities. However, the letters repeat (There are 2 Ns and 3 As), the actual # of possibilities would be 720/3!2!= 60

Then the number of possibilities would be 5!/2! because we combine the 3 As into one so there would be 5 "terms." Since the Ns repeat twice, we divide 5! by 2! to get 60 also.

See then we get the probability as 60/60, which can't be. What did I do wrong?

Answer 1

In the second case you have but $4$ characters, not $5$: $B, A^3, N_1, N_2$.

Answer 2

When you combine the 3 A's into 1, there are 4 resulting terms, not 5.

Thus the number of possible permutations where the 3 A's are together is $4!/2 = 12$, and the probability is $12/60 = 1/5$

Answer 3

I agree the total number of rearrangements of the word BANANA is $$ 6 \cdot \binom{5}{2} = \frac{6 \cdot 5 \cdot 4}{2!} = 60. $$

The actual possibilities are picked by fixing one of 4 places for the leading one of the AAA and picking one of the remaining $3$ places for the B, so there are a total of $4 \cdot 3 = 12$ ways to do that.