Problema Solution
Determine how many four-character codes can be formed if the first character is a nonzero digit, the second character is an odd digit, the third character is any digit, and the last character is a letter of the alphabet, and repetition of digits is allowed.
(A digit is 0, 1, 2,3,4,5,6,8, or 9.)
Answer provided by our tutors
since ist character is a non zero
hence no of ways it can be used 8
no of ways to write second = 4 ways
no of ways of third = 10
no of ways of foruth = 26
hence total number of ways = 8 * 4* 10 *26
= 8320 ways