Choosing letters at random from A,A,A,B,B,R,R,R We arrange letters A,A,A,B,B,R,R,R randomly. Find a probability that they will form sequence: RABARBAR.
I have an answer and it is: $$\frac{1}{{{8}\choose{3}}{{5}\choose{2}}{{3}\choose{3}} }$$
Is this a correct answer if so why is that?
 A: There are 3 $A$s and $8$ possible spots for them, so the probability that they will be in the correct spots is $\frac{1}{8\choose3}$. Given that this has happened, there are $5$ spots for the two $B$'s so the probability that they are in the correct spot is $\frac{1}{5\choose2}$. Given that both of these have happened, the $R$s must be in the correct spots, i.e. this event has probability $1=\frac{1}{3\choose3}$. Now combine these conditional probabilities and find that this event will happen with probability $$\frac{1}{{8\choose 3}{5\choose 2} {3\choose 3} }.$$
A: The total ways are $\frac{8!}{3!.3!.2!}$ so the probability of RABARBAR is $\frac{3!.3!.2!}{8!}$
A: I would prefer the solution as 3!3!2!/8!
Out of all 8! permutations, you can have any permutation of 3 A's, 3 R's and 2 B's.
A: The total number of permutations is $8!=40320$, and this repeats each unique sequence $3!3!2!=72$ times.
Therefore there are $\frac{40320}{72}=560$ different sequences, and so RABARBAR appears with probability $1/560\approx0.0017857$.
