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- How many different three-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 if repetition is allowed and one's position is filled with the first prime number?
How many different three-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 if repetition is allowed and one's position is filled with the first prime number?
160
250
600
400
Answers (1)
To calculate the number of different three-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, with repetition allowed and the one's position filled with the first prime number, we can proceed as follows:
First, we have to consider the prime number options for the one's position, which is limited to the digits 2, 3, 5, or 7.
For the hundreds position and the tens position, any of the ten available digits can be chosen (0, 1, 2, 3, 4, 5, 6, 7, 8, or 9).
Therefore, the number of different three-digit numbers that can be formed is obtained by multiplying the choices for each position:
Number of choices for the one's position = 4 (since there are 4 prime numbers available: 2, 3, 5, 7)
Number of choices for the hundreds position = 10 (since any of the ten digits can be chosen)
Number of choices for the tens position = 10 (again, any of the ten digits can be chosen)
Total number of different three-digit numbers = Number of choices for the one's position Number of choices for the hundreds position
Number of choices for the tens position
= 4 10
10
= 400
Therefore, there are 400 different three-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, with repetition allowed and the one's position filled with the first prime number.
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