How many different positive four-digit integers can be formed if the first digit must be 2, the last digit cannot be 0, and digits may be repeated?
a) 336
b) 900
c) 1000
d) 9000
e) 10000
one way for first digit, 10 different possiblities for 2nd and 3rd place, 9 ways for unit digit
1x10x10x9 = ?
To find the total number of different positive four-digit integers that can be formed based on the given conditions, we'll consider each position separately.
The first digit must be 2, so there is only one choice for this position.
For the second digit, any of the digits from 0 to 9 can be chosen. Since there are 10 possible choices, there are 10 options for the second digit.
For the third digit, any of the digits from 0 to 9 can be chosen, including the possibility of choosing 0. So there are 10 options for the third digit.
The fourth digit cannot be 0, but it can be any of the digits from 1 to 9. Therefore, there are 9 options for the fourth digit.
By multiplying the number of options for each position, we can find the total number of different positive four-digit integers that can be formed:
1 (choice for first digit) x 10 (choices for second digit) x 10 (choices for third digit) x 9 (choices for fourth digit) = 900.
Therefore, the correct answer is option b) 900.