Question 1178006: using all digits from 0 to 9, allowing for repetition, how many 3 digit numbers are possible that contain at least one 5?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Using all digits from 0 to 9, allowing for repetition, how many 3 digit numbers are possible
that contain at least one 5?
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Let U be the set of all three-digit integer positive numbers (an "universal set").
Let A be a subset in U consisting of all three-digit integer positive numbers having "5" as the left-most digit.
Let B be a subset in U consisting of all three-digit integer positive numbers having "5" in the second position.
Let C be a subset in U consisting of all three-digit integer positive numbers having "5" in the third position.
The set A contains 10*10 = 100 numbers;
The set B contains 9*10 = 90 numbers (zero is prohibited as the left-most digit);
The set C contains 9*10 = 90 numbers (zero is prohibited as the left-most digit).
The sets A and B have intersection comprised of all positive integer numbers of the form 55X. There are 10 numbers in this intersection.
The sets A and C have intersection comprised of all positive integer numbers of the form 5X5. There are 10 numbers in this intersection.
The sets B and C have intersection comprised of all positive integer numbers of the form X55. There are 9 numbers in this intersection.
The sets A, B and C have TRIPLE intersection comprised of ONE positive integer number 555. There is 1 number in this intersection.
Now, we want to determine the number of elements in the union (A U B U C).
Use the Inclusion-Exclusion formula
n(A U B U C) = n(A) + N(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C) =
= 100 + 90 + 90 - 10 - 10 - 9 + 1 = 252.
ANSWER. In all, there are 252 such numbers.
Solved.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
I suppose a math teacher somewhere wrote this problem....
It's sure a good thing that teacher isn't trying to teach English!!
"using all digits from 0 to 9, allowing for repetition, how many 3 digit numbers are possible that contain at least one 5?"
(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.
(2) The question asks for the number of 3-digit numbers that contain AT LEAST one 5. Therefore, the "allowing for repetition" is superfluous.
(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.
So the question is really simply "How many 3-digit numbers contain at least one 5?"
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Now that we have gotten past the absurd way the problem was stated, we can solve it....
I'll let you find the answer; here is a strategy:
(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.
Add the numbers you found in (1), (2), and (3).
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