Question

How many words or strings of 12 letters can be formed from the symbols $$a, a, a, a, b, b, b, b, b, b, b, b$$ provided that no two \(a\) 's can occur together?

Short answer

126 ways.

Step-by-step solution

Count Total Positions for 'a' and 'b'

There are 12 positions to fill with 'a' and 'b'. We need to use 4 'a's and 8 'b's given by the symbols: a, a, a, a, b, b, b, b, b, b, b, b.

Determine Constraints for 'a'

Ensure that no two 'a's can be together. To accomplish this, consider the 8 'b's as fixed and find suitable positions for the 4 'a's such that they are not adjacent.

Arrange 'b's

Position the 8 'b's first. This creates 9 potential gaps—one before each 'b', between each pair of 'b's and one after the last 'b' (represented as: _ b _ b _ b _ ... _ b _ ).

Choose Positions for 'a's

From the 9 potential gaps, select 4 to place the 'a's, ensuring no 'a' is adjacent to another 'a'. This can be done in \(\binom{9}{4}\) ways.

Calculate the Total Number of Combinations

Calculate the number of ways to arrange the 'a's among the gaps: \(\binom{9}{4} = 126\).

Verify and Conclude

Verify constraints are met (no two 'a's are adjacent) and ensure total counts add to 12 letters. Since each configuration respects the constraints, our solution is valid.

Key concepts

Combinatorial Constraints

When dealing with permutations and combinations, sometimes we face restrictions or constraints on how items can be arranged. In this problem, a key constraint is that no two 'a's can sit side by side in the sequence. To manage this constraint, we first arrange the 8 'b's out of the total 12 available positions. By placing these 'b's, we create potential slots or gaps where 'a's can be placed. This approach of using gaps is crucial to ensure no 'a's are adjacent. Think of it like lining up the 'b's and inserting an 'a' in the spaces between them. For 8 'b's, there are clear spots before, between, and after the 'b's. We end up with 9 such gaps, giving us precise spots to place each non-adjacent 'a'. The idea of forming gaps by fixing some positions helps us also satisfy the constraint throughout our process. Every step ensures that when the constraint is followed, the intended outcome (non-adjacent 'a's) is achieved.

Binomial Coefficient

The binomial coefficient is a mathematical way to determine the number of combinations of a particular set without considering order. In simpler terms, it helps us find out how many ways we can choose a certain number of items from a larger set. This principle is applied directly in our problem when we calculate how many ways we can choose gaps for the 4 'a's to sit in among the 9 available gaps created by arranging the 'b's. This is expressed as \( \binom{9}{4} \), which translates to choosing 4 gaps from a total of 9. The formula for the binomial coefficient is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]In our case, \( n = 9 \) and \( k = 4 \), resulting in:\[ \binom{9}{4} = \frac{9!}{4!(9-4)!} = 126 \]The calculation shows us there are 126 ways to choose these gaps, ensuring our 'a's are perfectly placed without violating any constraint.

Counting Permutations

Permutations involve arranging all the members of a set into some sequence or order. However, in problems with constraints, like ours, counting permutations demands a more strategic approach to ensure all conditions are honored. Here, we are asked to count permutations of the sequence with a specific restriction, where no two 'a's can be adjacent. By addressing the arrangement of 'b's first and focusing on the placement of 'a's second, we effectively manage the sequence constraint.The chosen method is simpler and more reliable than trying all possible combinations and filtering out ones with adjacent 'a's afterwards. It builds the correct solution from the ground up. The calculation \( \binom{9}{4} \) resolves directly to counting valid permutations adhering to our rules.Ultimately, this structured setup helps to streamline our work, ensuring the permutations we count are exactly the ones that meet our combinatorial requirements.