Can someone help me with this problem?
Determine the number of positive integers less than 10,000 that can be formed from the digits 1, 2, 3, and 4 if repetitions are allowed.
Thank you!
since there are 4 choices for each digit, and the numbers may be 1,2,3 or 4 digits long, you have
4^1 + 4^2 + 4^3 + 4^4 = (4^5-1)/(4-1) = 1023/3 = 341
oops. Since 0-length numbers are not allowed, we really have only 340 possibilities.
Sure, I can help you with that problem!
To determine the number of positive integers less than 10,000 that can be formed from the digits 1, 2, 3, and 4 with repetitions allowed, we'll need to break the problem into several steps.
Step 1: Determine the number of choices for each digit position.
Since repetitions are allowed, each digit position can be filled with any of the four digits (1, 2, 3, or 4), giving us 4 choices for each position.
Step 2: Count the total number of possible integers.
Since we are looking for integers less than 10,000, we need to consider the following cases for the first digit:
- For the thousands digit, we can choose any digit from 1 to 4, giving us 4 choices.
- For the hundreds digit, again, we can choose any digit from 1 to 4, giving us 4 choices.
- For the tens digit, we have 4 choices again.
- For the units digit, once more, we have 4 choices.
To find the total number of possible integers, we multiply the number of choices for each digit position: 4 choices for the thousands digit multiplied by 4 choices for the hundreds digit multiplied by 4 choices for the tens digit multiplied by 4 choices for the units digit. So, the total number of possible integers is 4 * 4 * 4 * 4 = 256.
Therefore, there are 256 positive integers less than 10,000 that can be formed from the digits 1, 2, 3, and 4 if repetitions are allowed.
I hope this explanation helps! Let me know if you have any further questions.