How many five position codes can be made under these circumstances

Question

how many five position codes can be made under these circumstances: The first position is a number, the second position is a vowel, the third position as a consonant, and the fourth position is a number greater than four, and the last position is A, T or K? (assume repetition is allowed)

Answer 1

Number of choices available for the first position is 10 (because there are 10 unique codes in the decimal number system 0,1,2,3,4,5,6,7,8,and 9)

Number of choices available for the second position is 5 (because there are 5 vowels in the English alphabet a, e, i, o and u.

Number of choices available for the third position is 21 (excluding the 5 vowels from the total 26 alphabets)

Number of choices available for the fourth position is 4 ( from the 10 codes , we remove the codes 0, 1, 2, 3 and 4)

Number of choices available for the last position is 3 (has to be A, or T, or K)

The number of 5 position codes which can be made under these circumstances will be equal to the product of all the number of choices available. I want you to calculate this product.

Given that repetition is allowed, it means that it is ok to have the same number in the first position and the fourth position; it is ok to have the vowel a in both the second position and the last position and it is ok for the letters T and k to repeat in the third position and the last position.

Answer 2

(10)(5)(21)(6)(3) = ?