Find the number of different arrangements of the letters in the word

COONABARABRAN

How many of those arrangements (i) start with the six vowels; (ii) contain the

four letters ‘A’ consecutively; (iii) contain no two consecutive letters ‘A’?

Answer:

Total number of arrangements: $\displaystyle \binom{13}{1,2,2,2,4}$ = 16216200

(i) Put As and Os at start $\displaystyle \binom{6}{2,4}$ = 15. Then number of arrangements of NNRRBBC is $\displaystyle \binom{7}{2,2,2}$ = 630. 15 * 630 = 9450

(ii) Treat AAAA as one item. That means there are 10 items to arrange - AAAA C O O N N B B R R. $\displaystyle \binom{10}{2,2,2,4}$ = 18900 ways.

(iii) _C_O_O_N_N_B_B_R_R_ ---> Put As in the gaps. This can be done $\displaystyle \binom{10}{4}$ = 210 ways. Number of arrangements of COONNBBRR = $\displaystyle \binom{9}{2,2,2,2}$ = 22680. 210 * 22680 = 4762800.

Are those answers correct?