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
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