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Combinatorics: Questions Involving Internal Orderings

Jerome randomly selects an arrangement of the letters in the word "EXCESS." What is the probability that the arrangement he selects will have a vowel for the first letter and a consonant for the last letter?
Incorrect. [[snippet]]
__Alternative Explanation__: Treat the problem as a series of events with no repetition, and figure out the probability for each event. Since the first and last digits have limitations, start there. The probability of the first letter being a vowel is $$\frac{2}{6}$$ (two vowels, E and E, out of six letters). The probability of the last letter being a consonant is now $$\frac{4}{5}$$ (four consonants—X, C, S, and S—out of five letters, as one was already chosen for the first). Once these conditions have been set, you don't really care what happens to the four letters between the first and last: for each of those, all of the remaining letters are wanted outcomes. Thus, the probability of a wanted outcome for the second letter is $$\frac{4}{4}$$ (four letters available, and all four are wanted outcomes), $$\frac{3}{3}$$ for the third letter, $$\frac{2}{2}$$ for the fourth, and $$\frac{1}{1}$$ for the fifth. In effect, the probability of a wanted outcome for the four middle letters is 1. Since we don't care what happens there, the probability of a wanted outcome is guaranteed. Thus, the final probability is >$$\text{Probability} = \frac{2}{6}\times1\times1\times1\times1\times\frac{4}{5} = \frac{8}{30} = \frac{4}{15}$$.
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Incorrect. [[snippet]]
Correct. [[snippet]] Let's start with the good choices. Start by filling the __SeBoxes__ according to the possible choices: * The first letter can only be a vowel. There are two vowels (the two E's), so there are two choices. * The sixth letter can only be a consonant, so there are four choices (X, C, S, and S). Then move to the other terms: * The second through fifth letters can be anything, but since we already used two letters, there are only four choices left for the second letter (since 2 out of 6 have already been used) and then three, two, and one choices, respectively, for the third, fourth, and fifth letters (the number decreases with each consecutive letter because there is no repetition). The __SeBoxes__ should therefore look like this: >$$\displaystyle\mathop{\fbox{${2}$}}\limits^{{1}{\text{st digit}}}\times \mathop{\fbox{${4}$}}\limits^{{2}{\text{nd digit}}}\times \mathop{\fbox{${3}$}}\limits^{{3}{\text{rd digit}}}\times \mathop{\fbox{${2}$}}\limits^{{4}{\text{th digit}}}\times \mathop{\fbox{${1}$}}\limits^{{5}{\text{th digit}}}\times \mathop{\fbox{${4}$}}\limits^{{6}{\text{th digit}}} $$.
But since we have two E's and two S's, this number of arrangements includes some redundant choices displacing similar letters. These duplicate letters turn this question into an internal-ordering problem. The internal ordering between the two E's and the two S's doesn't matter, so we need to divide twice by $$2!$$ so as to get rid of the redundant arrangements. Thus, the good arrangements are actually >$$\text{Good arrangements} = \require{cancel}\frac{2\times4\times3\times2\times1\times\cancel{4}}{\cancel{2!} \times \cancel{2!}} = 2\times4\times3\times2\times1 = 48$$ Out of how many combinations? There are $$6!$$ arrangements for 6 letters, but since there are two E's and two S's, divide twice by $$2!$$: >$$\text{Total arrangements} = \require{cancel}\frac{6!}{2!\times2!} = \frac{6\times5\times\cancel{4}\times3\times2\times1}{\cancel{2!} \times \cancel{2!}} = 6\times5\times3\times2=180$$. Thus, the fraction of good choices is >$$\text{Probability} = \frac{48}{180} = \frac{24}{90} = \frac{12}{45} = \frac{4}{15}$$.