Combinatorics: Questions Involving Internal Orderings

What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

__Alternative Solution__: You could also consider the probability of an E for the first letter _and_ an R for the last letter. * There are 13 letters and 3 E's, so the probability that the first letter is an E is $$\frac{3}{13}$$. * Once an E is the first letter, there are 12 remaining letters, 2 of which are R's. That means the probability the last letter is an R is $$\frac{2}{12}$$. Since there are no restrictions on the middle letters, you only need to multiply these two probabilities together to get the probability of getting an E first and an R last: >$$\text{Probability} = \frac{3}{13}\times \frac{2}{12} = \frac{1}{13} \times \frac{1}{2} = \frac{1}{26}$$.

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Incorrect. [[snippet]]

Correct. [[snippet]] Here are the total number of arrangements: There are 13 letters with some duplicates: 3 E's, 2 N's, 2 A's, and 2 R's. There are $$13!$$ arrangements for 13 different letters, but you have to divide by the number of internal arrangements of the four groups of duplicate letters (remember, there are $$3!$$ ways to arrange the 3 E's). >$$\text{Total} = \frac{13!}{3!2!2!2!}$$

Now for the good choices—where the first letter is E and the last one is R. Once an E is placed for the first letter and an R is placed in the last letter, there are 11 letters remaining, including 2 N's, 2 E's and 2 A's. There are $$11!$$ arrangements for 11 different letters, but you have to divide by the number of internal arrangements of the 3 groups of duplicate letters. >$$\text{Good combinations} = \frac{11!}{2!2!2!}$$ Thus, the probability is >$$\displaystyle \text{Probability} = \require{cancel}\frac{\frac{11!}{2!2!2!}}{\frac{13!}{3!2!2!2!}}$$ >>>$$= \frac{11!}{\cancel{2!}\cancel{2!}\cancel{2!}} \times \frac{3!\cancel{2!}\cancel{2!}\cancel{2!}}{13!}$$ >>>$$=\frac{3!}{13\times12}$$ >>>$$=\frac{6}{13\times12}$$ >>>$$=\frac{1}{26}$$

$$\frac{1}{13}$$

$$\frac{1}{20}$$

$$\frac{1}{26}$$

$$\frac{1}{50}$$

$$\frac{1}{100}$$

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