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John would make the 3-letter codes with 26 alphabets in cond

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by Max@Math Revolution » Tue Feb 16, 2016 12:57 am
John would make the 3-letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different each other with consonant. How many cases of the codes are there?

A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500


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by Max@Math Revolution » Thu Feb 18, 2016 5:51 pm
John would make the 3-letter codes with 26 alphabets in condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and both are consonant. How many cases of the codes are there?

A. 1,980
B. 2,020
C. 2,100
D. 2,200
E. 2,500


--> There should be a vowel in the middle of the 3-letter codes, which means 5 letters can be in the middle. Then, 21 letters can be in the first letter and 20 letters can be in the last letter as there should be different letters respectively. That is, 21*5*20=2,100.
Therefore, the answer is C.
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