Discrete Math distinct words. "Let m, n, and r be non-negative integers. How many distinct "words" are there consisting of m occurrences of the letter A, n occurrences of the letter B, r occurrences of the letter C, such that no subword of the form CC appears, and no other letters are used?"

kadejoset 2022-07-18 Answered
Discrete Math distinct words
"Let m, n, and r be non-negative integers. How many distinct "words" are there consisting of m occurrences of the letter A, n occurrences of the letter B, r occurrences of the letter C, such that no subword of the form CC appears, and no other letters are used?"
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Answers (2)

Jeroronryca
Answered 2022-07-19 Author has 13 answers
Step 1
Not the best possible answer, because it doesn't explain where your solution breaks down, but a different derivation...
Following the same basic principle, you want to know the number of words of the right form that have a CC in them. This has to start somewhere, and there are n + m + r 1 places it could start - i.e. the first C in the pair can occur as all but the last letter.
Step 2
Then you just need to fill up the remaining letters with n As, m Bs and r 2 Cs (so we should make some allowances for the cases r = 0 and r = 1, when we won't deduct anything at this stage anyway as there are no words with a CC subword).
Step 2
In the same way you found the total number of words originally, there are:
( n + m + r 2 ) ! n ! m ! ( r 2 ) !
ways of completing the word, having put a CC somewhere. So your answer should be:
( n + m + r ) ! n ! m ! r ! ( n + m + r 1 ) ! n ! m ! ( r 2 ) !
(providing r 2).
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Joanna Mueller
Answered 2022-07-20 Author has 5 answers
Step 1
There are ( m + n m ) words of length m + n that use m A's and n B's.
Any such word determines m + n + 1 "gaps" (including the 2 endgaps) into which we can slip a C.
Step 2
We can choose r of these gaps in ( m + n + 1 r ) ways. The number of words is therefore
( m + n m ) ( m + n + 1 r ) .
(By convention, ( a b ) = 0
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