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))#
