How many ways to rearrange 9 a's, 2 b's, and 2c's so that we never see bc and both b's always appear before both c's?
To find the number of ways to rearrange 9 a's, 2 b's, and 2 c's under the given conditions, we can use the concept of permutations with restrictions.
Let's break down the problem into steps:
Step 1: Calculate the total number of arrangements without any restrictions.
In this case, we have a total of 13 characters (9 a's, 2 b's, and 2 c's). The total number of arrangements without any restrictions can be found using the formula for permutations with repetitions:
Total arrangements = (13!) / (9! * 2! * 2!)
Step 2: Consider the arrangement where bc appears together.
To calculate this, we can treat bc as a single unit. So, we have 12 units to arrange (9 a's, 1 bc, 2 b's, and 2 c's). The total number of arrangements with bc appearing together can be found using the formula for permutations with repetitions:
Arrangements with bc together = (12!) / (9! * 2! * 2!)
Note: Since there are 2 b's and 2 c's, in the arrangements where bc appears together, b's and c's can be interchanged amongst themselves.
Step 3: Consider the arrangement where both b's appear before both c's.
To calculate this, we need to arrange the 9 a's, 2 b's, and 2 c's in such a way that both b's appear before both c's. We can treat b's as a single unit and c's as a single unit. So, we have 10 units to arrange (9 a's, 1 bb, and 1 cc).
Arrangements with both b's appearing before both c's = (10!) / (9! * 1! * 1!)
Step 4: Calculate the final result.
To find the number of ways to rearrange the characters such that we never see bc and both b's always appear before both c's, we need to exclude the cases where bc appears together. So, we subtract the arrangements with bc together from the arrangements with both b's appearing before both c's:
Number of ways = Arrangements with both b's appearing before both c's - Arrangements with bc together
Now, we can calculate the final result by substituting the values in the formula:
Number of ways = ((10!) / (9! * 1! * 1!)) - ((12!) / (9! * 2! * 2!))
Simplifying this expression will give us the answer.