1.) A security code has seven characters. Each code begins with two digits, then two letters, and finally three digits. the number of distinct security codes possible if repetition is allowed?
2.) How many ten digit phone numbers are possible if a number may not begin with a 0 or a 1?
216
1.) To find the number of distinct security codes possible with repetition allowed, we need to determine the number of choices for each character in the code.
- The first character has 10 possible choices since it can be any digit from 0 to 9.
- The second character also has 10 possible choices since repetition is allowed.
- The third and fourth characters are letters, so there are 26 choices for each character.
- The fifth, sixth, and seventh characters are digits, so each has 10 choices.
To find the total number of distinct security codes possible, we multiply the number of choices for each character together:
10 choices for the first character * 10 choices for the second character * 26 choices for the third character * 26 choices for the fourth character * 10 choices for the fifth character * 10 choices for the sixth character * 10 choices for the seventh character
Therefore, the total number of distinct security codes possible is:
10 * 10 * 26 * 26 * 10 * 10 * 10 = 6,760,000.
2.) To determine the number of ten-digit phone numbers possible, we need to consider the restrictions for the first digit.
Since the number may not begin with 0 or 1, we have 8 choices for the first digit (from 2 to 9). For the remaining nine digits, we have 10 choices for each digit (from 0 to 9) since repetition is allowed.
To find the total number of possible phone numbers, we multiply the number of choices for each digit together:
8 choices for the first digit * 10 choices for each of the remaining nine digits
Therefore, the total number of ten-digit phone numbers possible is:
8 * 10^9 = 8,000,000,000.