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.