how many different positive four-digit integers can be formed if the first digit must be 2, the last digit cannot be zero, and digits may be repeated?
I know the answer is 900 but I'm not sure how I would get the answer? thanks!
If the first digit must be 2, and digits can be repeated, you can forget about the leading digit and just look for every THREE-digit integer (INCLUDING leading zeros) whose last digit isn't zero. That's every integer between 000 and 999 (of which there are 1000), excluding the 10% of them that end in zero (of which there are 100) - which makes 900 in total.
Only 1 choice for the 1st spot, 10 choices for the 2nd, 10 choices for the 3rd and 9 choices for the 4th
1x10x10x9 = 900
To count the number of different positive four-digit integers that can be formed with these conditions, we can break down the problem step-by-step:
Step 1: The first digit must be 2.
Since the first digit must be 2, we have only one option for this digit - 2.
Step 2: The last digit cannot be zero.
Since the last digit cannot be zero, we have 9 options for this digit - 1, 2, 3, 4, 5, 6, 7, 8, 9.
Step 3: The remaining two digits can be any digit from 0 to 9.
For each of the remaining two spots, we have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) to choose from because digits can be repeated.
Step 4: Calculate the total number of possibilities.
To find the total number of possibilities, we need to multiply the number of options for each step together.
Total possibilities = 1 (Step 1) × 9 (Step 2) × 10 (Step 3) × 10 (Step 4)
Total possibilities = 900
Therefore, there are 900 different positive four-digit integers that can be formed with the given conditions.
To find the number of different positive four-digit integers that can be formed with the given conditions, we can break down the problem into four separate cases.
Case 1: The second digit can be any digit from 0 to 9.
Since the first digit is fixed at 2, there are 10 choices for the second digit (0 to 9). For the remaining two digits, any digit from 0 to 9 can be used, including repetition. So, there are a total of 10 choices for each of the remaining two digits. Hence, the total number of integers in this case is 10 * 10 * 10 = 1000.
Case 2: The second digit cannot be zero.
Since the requirement is for the first digit to be 2, there are no restrictions for choosing the second digit. So, there are 10 choices for the second digit (0 to 9). For the remaining two digits, any digit from 0 to 9 can be used, including repetition. So, there are a total of 10 choices for each of the remaining two digits. Hence, the total number of integers in this case is 10 * 10 * 10 = 1000.
However, we need to subtract the number of integers in which the last digit is zero.
Case 3: The second digit can be any digit from 0 to 9, and the last digit is zero.
Since the requirement is for the first digit to be 2, there are 10 choices for the second digit (0 to 9). For the third digit, any digit from 0 to 9 can be used, including repetition. But since the last digit must be zero, there is only one choice for the last digit. Hence, the total number of integers in this case is 10 * 10 * 1 = 100.
Case 4: The second digit cannot be zero, and the last digit is zero.
Since the requirement is for the first digit to be 2, there are no restrictions for choosing the second digit. So, there are 10 choices for the second digit (0 to 9). For the third digit, any digit from 0 to 9 can be used, including repetition. But since the last digit must be zero, there is only one choice for the last digit. Hence, the total number of integers in this case is 10 * 10 * 1 = 100.
To find the total number of different positive four-digit integers, we add up the number of integers in each case and subtract the over-counted integers (in Case 3 and Case 4):
Total = (Case 1 + Case 2) - (Case 3 + Case 4) = (1000 + 1000) - (100 + 100) = 900.
Therefore, the answer is 900 different positive four-digit integers that can be formed with the given conditions.