From the word EVERGREEN, 5 letters are chosen at random and arranged into a string of letters. What is the probability that this string is palindromic?
[Math] 5-letter strings using the letters in the word “EVERGREEN”
combinatoricspermutationsprobability
Related Solutions
Marc's answer is almost correct, but for one point; the numerator counts the occurrence of ABCDABCD twice. (As Graham Kemp pointed out. )
To elaborate: if I let $n(i)$ be the number of string patterns with ABCD starting at the $i$-th letter, then
$n(1) = n(2) = n(3) = n(4) = n(5) = 26^4$
(and $n(6) = n(7) = n(8) = 0$ )
but as ABCDABCD is counted in both $n(1)$ and $n(5)$, we must subtract by 1 to compensate.
So the end result would be
$$\frac{\left(\sum\limits_{i=1}^8{n(i)}\right)-1}{26^8}=\frac{5\times26^4-1}{26^8}$$
which is almost, but not quite, equal to the aforementioned answer. (Albeit probably close enough for realistic purposes...)
On the other hand, if we were looking for the expected value of the occurrence of ABCD, then we shouldn't need to subtract to compensate. (I think)
For longer strings, we would need to subtract duplicates, then add back to compensate for triplets, and so on, and the general formula I'm not sure how to write...
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.
Best Answer
in the case that the probability distribution is not uniform, one may find an elegant answer via the use of a tree diagram. Possibilities are: ERERE, EEXEE, ERXRE, REEER, REXER, where X is an element of the set {V,G,N}. Computing possibilities gives $\frac{1}{42}$
Edit: in the case of EEXEE, R may also be chosen. But the general solution is the same , as is the answer: simple tree diagram combined with multiplication principle.