AAPL wrote:Economist GMAT

What is the probability that out of the combinations that can be made using all the letters of the word EXCESS, Jerome will randomly pick a combination in which the first letter is a vowel and the last letter is a consonant?

A. \(\frac{96}{320}\)
B. \(\frac{24}{180}\)
C. \(\frac{33}{100}\)
D. \(\frac{48}{180}\)
E. \(\frac{96}{180}\)

OA D

Let's first calculate the total no. of words that can be formed out of the word EXCESS. We see that there are 6 letters in EXCESS; out of which there are 2 Ss and 2 Es.

So, the total no. of possible words = 6! / (2!*2!) = 180

Now let's come to how many words start with a vowel and end with a consonant.

We have only two vowels: 2 Es; thus, there are 4 consonants: 1 X, 1 C, and 2 Ss.

Case 1: Placing E as the first letter and S as the last letter.

We are left with 4 positions to be filled with 1 X, 1 C, 1 E and 1 S. The no. of ways to fill 4 places with 4 distinct letters = 4! = 24;

Case 2: Placing E as the first letter and one between X and C as the last letter.

The no. of ways to select one letter between X and C for the last place = 2

We are left with 4 positions to be filled with 1 X/C, 1 E and 2 S. The no. of ways to fill 4 places with 2 distinct and 2 same letters = 4!/2! = 12;

No. of ways = 2*12 = 24

Thus, the total no. of ways = 24 + 24 = 48

Thus, the required probability = 48/180

The correct answer: D

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GMAT Classes San Francisco | GMAT Tutoring Boston | GRE Prep Philadelphia | TOEFL Prep Classes DC | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.