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 $=\frac{\frac{7 !}{2 ! 2 !}}{83160}=\frac{1}{66}$