# Combinatorics QUESTION

• Oct 21st 2012, 03:18 PM
maximus101
Combinatorics QUESTION
Let $A = {a_1, a_2, . . . , a_k}$ be an alphabet, and let n_i denote the number of appearances of
letter a_i in a word. How many words of length n in the alphabet A are there for which
$k = 3, n = 10, n_1 = n_2 + n_3$ and $n_2$ is even?
• Oct 21st 2012, 03:25 PM
skeeter
Re: Combinatorics QUESTION
this is not calculus ... thread moved to Discrete Math
• Oct 21st 2012, 04:58 PM
johnsomeone
Re: Combinatorics QUESTION
Quote:

Originally Posted by maximus101
Let $A = {a_1, a_2, . . . , a_k}$ be an alphabet, and let n_i denote the number of appearances of
letter a_i in a word. How many words of length n in the alphabet A are there for which
$k = 3, n = 10, n_1 = n_2 + n_3$ and $n_2$ is even?

"let n_i denote the number of appearances of letter a_i in a word"? Which word? Are you saying that there's a single unknown-yet-fixed word in the alphabet A which defined the numbers $n_1, n_2, n_3,... , n_k ?$
OR, are you treating $n_1, n_2, n_3,... , n_k$ as k distinct functions functions from the domain of all possible finite words in A into the non-negative integers?
I would guess the later, but I'm not sure.

With that assumption:

$10 = n_1+n_2 + n_3 = (n_2+n_3) + n_2 + n_3 = 2(n_2 + n_3).$

$\text{Thus } n_3 = 5-n_2 \text{ and } n_1 = n_2 + n_3 = 5.$

$\text{Case }n_2 = 0:$
$\text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 0 = 5.$

$\text{Case }n_2 = 2:$
$\text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 2 = 3.$

$\text{Case }n_2 = 4:$
$\text{Thus } n_1 = 5, n_3 = 5 - n_2 = 5 - 4 = 1.$

$\text{So ANSWER } = \text{ number of ways} (n_1 = 5, n_2 = 0, n_3 = 5)$

$+ \text{number of ways}(n_1 = 5, n_2 = 2, n_3 = 3)$

$+ \text{number of ways}(n_1 = 5, n_2 = 4, n_3 = 1),$

$\text{and each term there is a clearer combinatorial problem.}$

$\text{So ANSWER } = \text{ number of distinguishable permutations of } "AAAAACCCCC"$

$+ \text{ number of distinguishable permutations of } "AAAAABBCCC"$

$+ \text{ number of distinguishable permutations of } "AAAAABBBBC".$

For the middle example: imagine there are 10 slots, numbered 1 through 10, that you need to fill with those letters to make your word. Choose 5 slots to fill with the A's, and then, from the 5 remaining slots, choose 2 to fill with the B's (Obviously the rest are then filled in with C's.) The number of ways to do that will be the number of distinguishable length 10 words using 5 A's, 2 B's and 3 C's.