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MathRevolution
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No of Vowels: 5 (A, E, I ,O ,U)

No of Consonants: 21 (26-5)

Now,

The Middle letter can be filled up in 5 ways.
The First letter can be filled in 21 ways ( As only consonants are allowed)

As one of the consonants is used in the first letter, the third letter can be filled up in 20 ways, so that the letter must be different from each other.

So total number of ways = 21 x 5 x 20 = 2100
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Bunuel: Please untag probability
MathRevolution
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
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BrushMyQuant
Bunuel: Please untag probability
MathRevolution
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
___________
Done. Thank you.
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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