What percent of 5 letter combinations that can be made from the letters of the word VERMONT and in which each letter can be used only once are the combinations in which the first letter is a vowel and the last letter is a consonant?
A. 20%
B. 23.8%
C. 25%
D. 30.2%
E. 35%
How will i solve this problem?
OA B
What percent of 5 letter combinations
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Well, there are two ways I can see to solve this problem. Both involve determining the maximum number of possibilities, and that is what I will start by doing.
VERMONT has 7 letters, but we are forming 5-letter combinations. So for the first letter, we have 7 choices. For the second, we have 6 (as we have already used one), for the next 5, and so on. Thus, the total number of possibilities is 7*6*5*4*3.
Now we can either find possibilities that do not fit the requirements and subtract them from the total, or we can find possibilities that do fit the requirements and add them up. I will be taking the latter approach, though there is nothing wrong with the former one.
Since the first letter must be a vowel, we have only two choices for that. Since the last letter must be a consonant, we have 5 choices for that. For the second, third, and fourth slots, we will have 5, 4, and 3 choices. So the total number of possibilities is 2*5*5*4*3.
Accordingly, the percentage will be (2*5*5*4*3) / (7*6*5*4*3). I trust at this point it becomes apparent why I did not bother to do the multiplication above. Many numbers will cancel, bringing us down to: (2*5) / (7*6), which can be further reduced to 5/21. This will be a bit less than 25 percent, so answer choice (B) stands out.
VERMONT has 7 letters, but we are forming 5-letter combinations. So for the first letter, we have 7 choices. For the second, we have 6 (as we have already used one), for the next 5, and so on. Thus, the total number of possibilities is 7*6*5*4*3.
Now we can either find possibilities that do not fit the requirements and subtract them from the total, or we can find possibilities that do fit the requirements and add them up. I will be taking the latter approach, though there is nothing wrong with the former one.
Since the first letter must be a vowel, we have only two choices for that. Since the last letter must be a consonant, we have 5 choices for that. For the second, third, and fourth slots, we will have 5, 4, and 3 choices. So the total number of possibilities is 2*5*5*4*3.
Accordingly, the percentage will be (2*5*5*4*3) / (7*6*5*4*3). I trust at this point it becomes apparent why I did not bother to do the multiplication above. Many numbers will cancel, bringing us down to: (2*5) / (7*6), which can be further reduced to 5/21. This will be a bit less than 25 percent, so answer choice (B) stands out.
Elias Latour
Verbal Specialist @ ApexGMAT
blog.apexgmat.com
+1 (646) 736-7622
Verbal Specialist @ ApexGMAT
blog.apexgmat.com
+1 (646) 736-7622