5-letter strings using the letters in the word "EVERGREEN"

Question

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?

Answer 1

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.

Answer 2

The answer depends on the randomizing process. I'll assume the process is: pick one letter at random, add it to the 1st position, pick another letter at random, add it to the 2nd position, and so on. So e.g. a string like EEEEN could arise in $4!$ ways.

Here's a sketch of an approach.

Via this process, in total, there are $9 \times 8 \times 7 \times 6 \times 5$ 5-letter strings in total that can be formed. Now we want to know how many of these are palindromes (read the same back to front).

Except for the central letter, each other letter must occur in pairs. There's not many possibilities for the structure of these strings. E.g. they may look like ER*RE where * is a wildcard (it can be any remaining letter). [There's other possibilities.]

So we find all the structures, and for each of them compute the number of ways that each of these structures can arise. We then sum them up and divide by $\binom{9}{5} 5!$ to find the probability.

Answer 3

The word has 4 E's, 2 R's, 1 V, 1 G and 1 N

For the total sample space $\Omega$, consider all possible 5-letter strings, taking letters from the sets as described: $$ \left\{E,V,G,R,N\right\}: {}^5\!P_5 \\ 2E + \left\{ R,V,G,N\right\}: {}^4\!C_3 \cdot \frac{5!}{2!} \\ 3E + \left\{ R,V,G,N\right\}: {}^4\!C_2 \cdot \frac{5!}{3!} \\ 4E + \left\{ R,V,G,N\right\}: {}^4\!C_1 \cdot \frac{5!}{4!} \\ 2R + \left\{ E,V,G,N\right\}: {}^4\!C_3 \cdot \frac{5!}{2!} \\ 2R + 2E + \left\{ V,G,N\right\}: {}^3\!C_1 \cdot \frac{5!}{2! 2!} \\ 2R + 3E: \frac{5!}{2!3!} \\ \therefore |\Omega| = 840 $$ For the subset of palindromes S, the word is structured $X_f C X_b $, where X is an ordered string, $X_f$ is the string forward, and $X_b$ the string backward, and C a character in the middle, e.g. when X = "ER". From this one gets the following cases:

• X = "ER", "RE"; C = "V", "G", "E", "N" (e.g. when X = "ER" the string can be ERVRE)
• X = "EE"; C = "V", "G", "N", "R"

In total there are 12 such cases. So the probability is $\frac{12}{840} = \frac{1}{70}$.