1. ## Maddening Unexplained Conditional Permutations

Hey, the textbook decided to throw out some problems that it hadn't explained in the lesson on conditional permutations. I've been trying to figure it out myself, but i could really use some help.

Alright, it says "How many positive integers are there less than 900 such that all digits are odd?"

It explained how to do like three-digit numbers, but not individual integers.. Help would be totally appreciated!

2. There are five odd digits: {1,3,5,7,9}.
But 9 cannot be the first digit. So there are (4)(5)(5)=100.

3. Thanks Plato, but there's where i got my hangup. that solves for only three-digit numbers. not all integers.

actually, i think i just figured it out.

only using odd numbers so 5 is the greatest number of variables.

under 900 so 4 possible numbers the first digit.

4 x 5 x 5 = 100 (first possible outcome) then add another row for the two digit possible outcomes.
5 x 5 = 25 repeat for single digit...
5 = 5

100+25+5 = 130. 130 odd integers lower than 900.

eh?