Answer:
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 $1 \times 28 = 28$.