How many different four-digit numbers can be formed using the prime numbers less than 15, if repetition is allowed and ten's position is filled with the perfect square of 3?
212
216
214
210
To calculate the number of different four-digit numbers that can be formed using the prime numbers less than 15, with repetition allowed and the ten's position filled with the perfect square of 3, we can proceed as follows:
The perfect square of 3 is 9, so for the ten's position, we have only one option, which is 9.
For the thousands position, we can choose any prime number less than 15. The prime numbers less than 15 are 2, 3, 5, 7, 11, and 13. Since repetition is allowed, we have six options.
For the hundreds position, we can again choose any prime number less than 15. With repetition allowed, we have six options.
For the one's position, we can choose any prime number less than 15. With repetition allowed, we have six options.
Therefore, the number of different four-digit numbers that can be formed is obtained by multiplying the choices for each position:
Number of choices for the ten's position = 1 (since it is fixed as 9)
Number of choices for the thousands position = 6 (any of the six prime numbers less than 15)
Number of choices for the hundreds position = 6 (any of the six prime numbers less than 15)
Number of choices for the one's position = 6 (any of the six prime numbers less than 15)
Total number of different four-digit numbers = Number of choices for the ten's position Number of choices for the thousands position
Number of choices for the hundreds position
Number of choices for the one's position
= 1 6
6
6
= 216
Therefore, there are 216 different four-digit numbers that can be formed using the prime numbers less than 15, with repetition allowed and the ten's position filled with the perfect square of 3 (9).
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