Permutation and Combinations problem based on letter selection from a word

combinationscombinatoricspermutationssolution-verification

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.

Best Answer

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.

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