Question

consider the words consisting of 5 letter "a"'s, 7 letter "b"'s, and 1 letter "c". find the number of such words where there are no two consecutive "b"'s.

Answer 1

To avoid consecutive b's, the string of b's must be divided into one or more groups of 1, 2, or 3 b's. There are a total of 5 + 1 = 6 possible places to insert the dividers to create these groups. For example, suppose we have a string of 7 b's and we insert dividers between the 3rd and 4th b's, and between the 5th and 6th b's. This would create the following groups of b's: bb|bb|b. In general, the number of possible ways to divide the string of b's into groups of 1, 2, or 3 b's is equal to the number of ways to insert 6 dividers into a string of 7 b's. This is equal to the number of ways to choose 6 of the 8 possible places to insert the dividers, which is equal to $\binom{8}{6} = 28$.

Since the 5 a's and the 1 c must be arranged in a specific order, there is only 1 way to arrange these letters. Therefore, the total number of words consisting of 5 letter a's, 7 letter b's, and 1 letter c is equal to $1 \times 28 = 28$.

Answer 2

• 6 possible words.

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According to the question there are 7 'b' s and we can't have them together. We assume more than two consecutive 'b' s contain two consecutive 'b' s and hence not considering any option of having 3, 4, 6 or 7 'b' s.

There is only one way to place 'b' s apart which leaves us with 6 gaps.

There are 5 + 1 = 6 other letters which is just the number to fill in the gaps between 7 'b' s.

We have only an option for 'c' to have any of the 6 gaps, it gives us 6 options and no other options for 'a' s.

Therefore in total we have 6 possible ways.