When performing a trick a magician says the word ABRACADABRA. The 11 letters of this word are arranged in a row.

(a) Find the number of different arrangements that can be made.

[2]

(b) Find the number of different arrangements in which the $2 \mathrm{~B}$ ‘s are next to each other, the $2 \mathrm{R}$ ‘s are next to each other, exactly 4 of the A’s are next to each other, and the C is next to the D.

[3]

(c) Given that the 11 letters are arranged randomly, find the probability that all $5 \mathrm{~A}$ ‘s are together.

[3]

(a) $83160$ ;

(b) $144$ ;

(c) $\frac{1}{66}$

[Maximum marks: 8 marks]

(a) Number of ways $=\frac{11 !}{5 ! 2 ! 2 !}=83160$

(b) Number of ways $=3 ! \times 2 ! \times\left(\begin{array}{l}4 \\ 2\end{array}\right) \times 2 !=144$

(c) Required probability $=\frac{\frac{7 !}{2 ! 2 !}}{83160}=\frac{1}{66}$