In how many ways can the letters in the word "PROBABILITY" be arranged using the following restrcitions

Question

In how many ways can the letters in the word "PROBABILITY" be arranged if the first letter must be "B" and the last letter cannot be an "O", "A", or "I"?

Answer 1

Solved 2 x 6 x (9!) = 4,354,560. My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning

Answer 2

The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $\binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is $$1 \cdot 6 \cdot \binom{9}{2} \cdot 7! = 6 \cdot \frac{9!}{2!7!} \cdot 7! = 3 \cdot 9! = 1,088,640$$

Answer 3

first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $\dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)

so, total number of ways $=1\times 6\times\dfrac{9!}{2!}=1088640 $ ways