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  • How many different three-digit numbers can be formed using the digits 1, 3, 4, 5, 6, 7, 8, and 9 if repetition is not allowed and one's position is filled with the first prime number? <p

How many different three-digit numbers can be formed using the digits 1, 3, 4, 5, 6, 7, 8, and 9 if repetition is not allowed and one's position is filled with the first prime number?

 

Option: 1

42


Option: 2

26


Option: 3

34


Option: 4

45


Answers (1)

best_answer

To calculate the number of different three-digit numbers that can be formed using the digits 1, 3, 4, 5, 6, 7, 8, and 9, with repetition not allowed and the one's position filled with the first prime number, we can proceed as follows:

For the one's position, we have only one option, which is the first prime number, 2.

For the hundreds position, there are 7 remaining digits to choose from, as we have used one digit already.

For the tens position, there are 6 remaining digits to choose from after using two digits.

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 = 1 (since it is fixed as 2)

Number of choices for the hundreds position = 7 (as there are 7 remaining digits to choose from)

Number of choices for the tens position = 6 (after using two digits, 6 digits remain)

Total number of different three-digit numbers = Number of choices for the one's position \times Number of choices for the hundreds position \times Number of choices for the tens position

= 1 \times 7 \times 6

= 42

Therefore, there are 42 different three-digit numbers that can be formed using the digits 1, 3, 4, 5, 6, 7, 8, and 9, with repetition not allowed and the one's position filled with the first prime number (2).

 

Posted by

Divya Prakash Singh

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