let the color code be ABC
we have 5 vowels a,e,i,o,u
B cab have any of the 5 vowels.
So,there are 5 ways
21 consonants
there are 21 ways to fill A.
as digits cannot be repeated.therefore,we can fill C in 20 ways
Totalcodes possible=21*5*20
=2100

+1 Kudo,If i helped
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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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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

Bunuel: Please untag probability
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Done. Thank you.
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