There are ((9),(2)) ways to place the "B"s.
From the remaining 7 spots, there are ((7),(3)) ways to place the "A"s.
From the remaining 4 spots, there are ((4),(2)) ways to place the "R"s.
In the remaining 2 spots, there are 2! ways to arrange the "I" and the "N"
Thus the total possible arrangements is
((9),(2))((7),(3))((4),(2))2! = (9!)/(7!2!)(7!)/(4!3!)(4!)/(2!2!)2! =(9!)/(3!2!2!) = 15120
Note that we could also write this as
(9!)/(3!2!2!1!1!)
with the factorials in the denominator matching the "groups" of letters (3 "A"s, 2 "B"s, 2 "R"s, 1 "I", 1 "N"). Looking at how the cancellation worked when we did our initial multiplication, we can see how it will work out this way in general, and so as a shorthand, we can write this using the multinomial coeffficient
((,,9,,), (3,2,2,1,1))
