# Arrangements of letters in COONABARABRAN

• Dec 11th 2011, 01:15 PM
nukenuts
Arrangements of letters in COONABARABRAN
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’?

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

(i) Put As and Os at start $\binom{6}{2,4}$ = 15. Then number of arrangements of NNRRBBC is $\binom{7}{2,2,2}$ = 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. $\binom{10}{2,2,2,4}$ = 18900 ways.

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

• Dec 11th 2011, 01:28 PM
Plato
Re: Arrangements of letters in COONABARABRAN
Quote:

Originally Posted by nukenuts
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’?
(ii) Treat AAAA as one item. That means there are 10 items to arrange - AAAA C O O N N B B R R. $\binom{10}{2,2,2,{\color{red}4}}$ = 18900 ways.

In (ii) there is no need for the 4, "AAAA" is only one.
• Dec 11th 2011, 01:45 PM
nukenuts
Re: Arrangements of letters in COONABARABRAN
Quote:

Originally Posted by Plato
In (ii) there is no need for the 4, "AAAA" is only one.

Oops, meant to treat it as one item but wrote it down wrong.