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