compute the number of sequences formed from 3 a's, 5 b's, and 8 c's. a) that have no consecutive b's b) that have no consecutive symbols the same
compute the number of sequences formed from 3 a's, 5 b's, and 8 c's. a) that have no consecutive b's b) that have no consecutive symbols the same
Chapter9: Sequences, Probability And Counting Theory
Section9.5: Counting Principles
Problem 36SE: The number of 5-element subsets from a set containing n elements is equal to the number of 6-element...
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Question
compute the number of sequences formed from 3 a's, 5 b's, and 8 c's.
a) that have no consecutive b's
b) that have no consecutive symbols the same
The textbook solution is a) C(12,5)11!/(8!3!) b) 2C(8,3)+14C(6,4)
I think I understand how to get a, I am just a bit confused on where the 11!/(8!3!)) comes from exactly. I am super lost on b. Thanks!
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