Question 1178006
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I suppose a math teacher somewhere wrote this problem....<br>
It's sure a good thing that teacher isn't trying to teach English!!<br>
"using all digits from 0 to 9, allowing for repetition, how many 3 digit numbers are possible that contain at least one 5?"<br>
(1) The digits (in our familiar base 10 number system) are 0 through 9; there are 10 of them.  There are clearly NO 3-digit numbers that use all of those digits ("using all digits from 0 to 9").  If the numbers are to be 3-digit numbers, then that opening phrase is not only unnecessary but contradictory; it does not belong in the statement of the problem.<br>
(2) The question asks for the number of 3-digit numbers that contain AT LEAST one 5.  Therefore, the "allowing for repetition" is superfluous.<br>
(3) What's left after we get rid of the unnecessary first two phrases is "how many 3 digit numbers are possible that contain at least one 5?"  In that remaining phrase, the "... are possible that" is unnecessary.<br>
So the question is really simply "How many 3-digit numbers contain at least one 5?"<br>
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Now that we have gotten past the absurd way the problem was stated, we can solve it....<br>
I'll let you find the answer; here is a strategy:<br>
(1) Count how many numbers have a 5 in the hundreds place.
(2) Count how many numbers have a 5 in the tens place -- but don't count the ones with a 5 in the hundreds place also, because you have already counted those.
(3) Count how many numbers have a 5 in the ones place -- but don't count the ones that also have a 5 in either the hundreds place or the tens place, because you have already counted those.<br>
Add the numbers you found in (1), (2), and (3).<br>