Permutation and Combinations problem based on letter selection from a word I just want to check whether my solutions are correct. I don't have the answers.

Given the word TOMORROW (8 letters),
(i) in how many ways can the word be arranged if the two R's are each at one end and the O's are not all together?
(ii) four letters are chosen at random. calculate the probability that the four letters chosen consist of at least 1 O AND at least 1 R.
(iii) four letters are chosen at random. calculate the probability that the four letters chosen consist of at least 1 O OR at least 1 R.

So, 8 letters with the following reps;
R, R
O, O, O
Now, part (i) asked for a case where the O's are not all together which means that 3 O's aren't allowed to be together but 2 O's are allowed to be together, right?
Here's my solution to part (i);
first, I calculate the case where the only restriction is the two R's at each end
R_{6 letters in the middle with 3 repeating O's}_R
this will give $\dfrac{6!}{3!} = 120$
then, I calculate the case where the O's are together along with the two R's at each end
R(OOO){3 letters}R
I'm going to count the group of O's as one letter here
this will give $4! = 24$
finally, I will subtract $24$ from $120$ which gives $96$
So my answer to part (i) is $96$.
However, I am not so sure if "not all together" means what I mentioned above. Does it mean that no O's should be next to each other? Someone please help me on that.
Here's my solution to part (ii);
So, $4$ letters are to be chosen from the $8$ letters in the word TOMORROW.
I will have to be choosing letters from these only now; TMW ($3$ letters)
When the restrictions are applied, I guess the following are the only possible combinations;
O O O R = 1
O O R R = 1
O R R _ = 3C1 = 3
O R _ _ = 3C2 = 3

Total gives $8$
Now for the probability, this will give $$\frac{8}{\dbinom{8}{4}} = \frac{8}{70}$$
Hence, my answer to part (ii) is $8/70$.
Here's my solution to part (iii);
again, choosing from only these 3 letters now; TMW
O O O _ = 3C1 = 3
O O _ _ = 3C2 = 3
O _ _ _ = 3C3 = 1

R R _ _ = 3C2 = 3
R _ _ _ = 3C3 = 1

Total gives $11$.
So, my answer to part (iii) is $11/70$.
Please help. Thanks in advance.
 A: For part (i), it says that O's are not all together. So I agree with your interpretation and the working.
Part (ii) asks for probability that there is at least one O and at least one R.
So we first find probability of all $ \small 4$ letter selections where
a) there is no O  = $ \displaystyle \small {5 \choose 4} / {8 \choose 4} = \frac{5}{70}$
b) there is no R  = $ \displaystyle \small {6 \choose 4} / {8 \choose 4} = \frac{15}{70}$
c) there is no O and no R - zero as no such selection is possible
So desired probability $ \displaystyle \small  = 1 - \frac{5}{70} - \frac{15}{70} = \frac{5}{7}$
Part (iii) asks for probability that there is at least one O or at least one R.
Based on interpretation that we also count cases where there are both O and R, this is certain. If we leave O and R, there are only $3$ letters T, M , W. But as we have to choose $4$ letters, there will be at least one O or one R.
But if the question means that at least one R or at least one O, not both then it is addition of (a) and (b) in part (ii) as (a) has no O but has at least one R. (b) has no R but has at least one O.
A: Your solution seems almost correct, but there are a few counting issues:
Fast solution
i) not all together means that $< 3$ together is allowed
ii)+iii) Instead of writing down all possible allowed combinations, you can use the counter probability as well. Imagining taking 100 letters from 1000 possible characters, listing all combinations would be impossible.
As you see, 5 of 8 characters are Os or Rs, that means choosing 4 of those 8 characters will always include an O or an R $\Rightarrow$ $P(\text{at least 1 "O" or 1 "R"}) = 1$.
ii) $P(\text{at least one O AND one R}) = 1 - P(\text{No O (but R allowed)}) - P(\text{No R (but O allowed)}) - P(\text{Neither O nor R}) = 1 - \binom{5}{4}/70 - \binom{6}{4}/70 - 0 = \frac{20}{70}$
The probabilities are calculated as follows: If no "O" is allowed only the other $5$ letters are allowed, if no "R" is allowed only the other $6$ letters can be taken, therefore you need the two binomial coefficients $\binom{5}{4}$ and $\binom{6}{4}$.
iii) The word 'OR' is normally meant to be a non-exclusive or, meaning that $\geq 1$ 'O' AND $\geq 1$ 'R' are allowed as well. However, you are missing this type of letters (e.g. OOOR). The answer would then be $1$.
If "exclusive or" is meant, you can just use ii) to calculate the probability:
$P(\text{at least one O XOR at least one R}) = 1 - P(\text{at least one O AND at least one R}) - P(\text{no O AND no R}) = 1 - ii) - 0 = \frac{20}{70}$
Counting correctly
Now let's come to your mistakes while counting:
As you're using the binomial coefficient - meaning you are using distinguishable objects - you can't count the option "OORR" only once. In fact there are $\binom{3}{2}$ possibilities to choose 2 of those 3 Os. So your real possibilities for ii) are:
OOOR: 2C1 = 2
OORR: 3C2 = 3
OOR_: 3C2*2C1*3 = 18
ORR_: 3C1*3 = 9
OR__: 3C1*2C1*3C2 = 18

Total of $50$, $P=\frac{50}{70}$
And for the next part iii)
OOO_ = 3C1 = 3
OO__ = 3C2 * 3C2 = 9
O___ = 3C1 * 3C3 = 3
RR__ = 3C2 = 3
R___ = 2C1 * 3C3 = 2

Total of $20$.
