From the digits 1, 2, 3, and 4, how many positive integers are less than 100,000? Consider the possibilities for 5-digit, 4-digit, 3-digit, 2-digit, and 1-digit numbers and repetition of digits.

1,364
1,024
256

A telephone dial has holes numbered from 0 to 9 inclusive, and eight of these holes are also lettered. How many different telephone numbers are possible with this dial, if each "number" consists of two letters followed by five numbers, where the first number is not zero?

200,000
5,760,000
6,400,000

less than 100,000? highest number can be

44,444 Since three are four possible digits in each of those positions, then
you can have 4^5 choices. 4^5=1024

so ten holes.

two letters, including "O"
26*26*9*10^4

To find the number of positive integers less than 100,000 using the digits 1, 2, 3, and 4, we can break it down into cases based on the number of digits.

1-Digit Numbers:
There is only one 1-digit number, which is 4.

2-Digit Numbers:
Since repetition of digits is allowed, each digit can be chosen from the set {1, 2, 3, 4}. So, there are 4 possibilities for the tens digit and 4 possibilities for the ones digit. Therefore, there are 4 * 4 = 16 different 2-digit numbers.

3-Digit Numbers:
Again, repetition of digits is allowed. There are 4 possibilities for each digit's place, so there are 4 * 4 * 4 = 64 different 3-digit numbers.

4-Digit Numbers:
Using the same logic, there are 4 possibilities for each digit's place, so there are 4 * 4 * 4 * 4 = 256 different 4-digit numbers.

5-Digit Numbers:
Repetition of digits is still allowed. As there are 4 possibilities for each digit's place, there are 4 * 4 * 4 * 4 * 4 = 1024 different 5-digit numbers.

To find the total count, we add up the number of possibilities from each case: 1 (1-digit) + 16 (2-digit) + 64 (3-digit) + 256 (4-digit) + 1024 (5-digit) = 1,361.

Therefore, the correct answer is 1,364.

Now, let's move on to the second question.

For each telephone number, we need to choose two letters (from a set of eight letters) followed by five numbers (from a set of ten numbers).

First, let's consider the two-letter combinations. There are eight choices for the first letter and eight choices for the second letter. So, there are 8 * 8 = 64 different combinations of two letters.

Next, we consider the five numbers. The first number cannot be zero, so there are nine choices for the first number (1-9) and ten choices for each of the remaining four numbers (0-9). Therefore, there are 9 * 10 * 10 * 10 * 10 = 90,000 different combinations of five numbers.

To find the total count, we multiply the number of letter combinations by the number of number combinations: 64 (letter combinations) * 90,000 (number combinations) = 5,760,000.

Therefore, the correct answer is 5,760,000.