• Nov 6th 2008, 10:34 AM
Savior_Self
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!
• Nov 6th 2008, 10:46 AM
Plato
There are five odd digits: {1,3,5,7,9}.
But 9 cannot be the first digit. So there are (4)(5)(5)=100.
• Nov 6th 2008, 10:51 AM
Savior_Self
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?