How many 3 digit integers can be chosen such that none of the digits appear more than twice, none of the digits equal 0?
To find the number of 3-digit integers where none of the digits appear more than twice and none of the digits are equal to 0, we can break it down into cases:
Case 1: The three digits are all different.
For the first digit, you have 9 choices (1 to 9), for the second digit, you have 8 choices (excluding the digit you used for the first digit), and for the third digit, you have 7 choices (excluding the two digits already used). So, in this case, you have 9 * 8 * 7 = 504 possibilities.
Case 2: Two of the three digits are the same.
There are three possible positions for the two identical digits: AABB, ABBA, and BAAB.
For the AABB pattern:
- You have 9 choices for A (1 to 9).
- You have 8 choices for B (excluding the digit used for A).
So, for this pattern, you have 9 * 8 = 72 possibilities.
Now, for the ABBA and BAAB patterns:
- You have 9 choices for A (1 to 9).
- You have 8 choices for B (excluding the digit used for A).
For each of these patterns, you have 2 possibilities (ABBA and BAAB).
So, for both patterns combined, you have 2 * 9 * 8 = 144 possibilities.
Now, add the possibilities from all cases:
Total possibilities = 504 (Case 1) + 72 (AABB) + 144 (ABBA/BAAB) = 720
So, there are 720 3-digit integers that meet the given criteria.
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