How do I find the total number of arrangements with multiple restrictions?

Question

Q) Find the total number of different arrangements of the 8 letters in the word TOMORROW that have an R at the beginning and an R at the end, and in which the three Os are not all together.

All I can start with here is the different ways to arrange the other letters in among the Os in 3! ways and times 2 for the Rs however can’t come up with much more beyond that. Any help and explanation would be greatly appreciated :)

Answer 1

Given that the R's must be at the beginning and the end the question can be simplified into the number of ways the letters TOMOOW can be arranged between the 2 R's without the three O's being together.

See the layout below:

R __ __ __ __ __ __ R

The number of different ways those 6 Letters can be arranged is 6!/(3!) = 120 ways.

Note that the permutation of 6 letters is divided by 3 factorial for 3 repeating letters.

Then you must subtract the number of ways where all 3 O's are consecutive (see belows)

R O O O __ __ __ R

R __ O O O __ __ R

R __ __ O O O __ R

R __ __ __ O O O R

For each of these 4 cases the other 3 letters can be arrange 3! or 6 ways. Thus there are 4*6=24 ways where the 3 O's are consecutive.

Thus the total number of different ways to arrange the letters in the word TOMORROW where the arrangement starts and ends with an R and does not have 3 consecutive O's is 120-24= 96 ways!