How many ways can the letters of the word TOMORROW be arranged if the Os can't be together? How many ways can the letters of the word TOMORROW be arranged if the Os cant be together?
I know TOMORROW can be arranged in $\frac{8!}{3!2!} = 3360$ ways. But how many ways can it be arranged if the Os can't be together? And what is the intuition behind this process?
 A: We interpret the question as saying we cannot have two  (or three) O's together. Think of the slots occupied for the remaining $5$ letters.  There are $6$ spaces "between" these slots for the O's to be squeezed into, no more than one O per space. Here the number of spaces is $6$ because I am counting the two endspaces. 
We choose $3$ of these $6$ spaces for the O's. This can be done in $\dbinom{6}{3}$ ways.
For each of these ways, the T can be placed in $5$ ways, then the M in $4$ ways, then the W in $3$ ways. Now it is all done, the R's take the remaining two slots. So our count is
$$\binom{6}{3}(5)(4)(3).$$ 
Remark: There are many other ways of counting. The advantage of this one is that it generalizes smoothly to a situation where the length of the word, and the number of O's, is much larger. 
The idea can be adapted for similar problems. A standard one is to ask how many ways can we line up $9$ adults and $5$ children in a row if no two childen can be next to each other.
A: First, you have to remove the permutations like this TOMOORRW and TMOOORRW, so see OO as an element, then we have $3360-\frac{7!}{2!}$.
EDITED
Now, Notice you remove words like this TOMOORRW and TMOOORRW with OO and OOO, but you remove a little bit more you want, why? can you see this? how to fix this problem?
