One non-zero digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?

The password consists of three characters: one non-zero digit, one vowel, and one consonant. The constraints are:

1. The password must start with a vowel.
2. The password must end with a consonant.

Let's determine the number of choices for each type of character:

• Number of non-zero digits: Digits are 1, 2, 3, 4, 5, 6, 7, 8, 9. So, there are 9 choices.
• Number of vowels (capital): A, E, I, O, U. So, there are 5 choices.
• Number of consonants (capital): Total letters = 26. Vowels = 5. Consonants = 26 - 5 = 21 choices.

Now, let's place these characters according to the given rules:

• The first position must be a vowel. Number of choices for the first position = 5 (any of A, E, I, O, U).
• The third (last) position must be a consonant. Number of choices for the third position = 21 (any of the 21 consonants).
• The remaining character is the non-zero digit, which must occupy the middle (second) position. Number of choices for the second position = 9 (any of the 9 non-zero digits).

To find the total number of such passwords, we multiply the number of choices for each position: Total passwords = (Choices for 1st position) * (Choices for 2nd position) * (Choices for 3rd position) Total passwords = 5 * 9 * 21 Total passwords = 45 * 21 Total passwords = 945

Therefore, 945 such passwords can be generated.

The final answer is C) 945.