Question 567696: how many numbers between 100 and 1000 have at least one 2 in the number
Found 2 solutions by Edwin McCravy, richard1234:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
There are these 7 cases, where the letters X and Y represent
numbers which are NOT 2's. Remember there are 9 digits which are
not 2.
The cases are 2XY, X2Y, XY2, X22, 22X, 2X2, 222
Case 1: 2XY
There are 9 ways to choose a digit for the X and for each of
those 9 ways there are 9 ways to choose a digit for the Y.
That's 9·9 or 81 ways
Case 2: X2Y
There are 8 ways to choose the X (it can't be 0 or 2) and for each
of those 8 ways there are 9 ways to choose the Y.
That's 8·9 or 72 ways
Case 3: XY2
That's just like Case 2. That's 72 more ways.
Case 4: X22
There are 8 ways to choose the X (it can't be 2 or 0). That's 8 more ways.
Case 5: 22X
There are 9 ways to choose the X. That's 9 more ways.
Case 6: 2X2
That's just like Case 5. That's 9 more ways.
Case 7: 222. That's 1 more way, simply the number 222.
81 + 72 + 72 + 8 + 9 + 9 + 1 = 252
Answer: 252
Edwin
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! There are 901 numbers between 100 and 1000 (inclusive). Instead of counting how many numbers have at least one 2, we can count how many numbers do not have any 2's at all, and subtract from 901 (since this is the complement of what we want).
For each of the three digits, there are 8 choices for the first digit (anything except 0, 2), 9 choices for the second digit (anything except 2) and 9 choices for the third digit. Hence the number of 3-digit numbers with no 2's is 8*9*9 = 648. Including the number "1000" we have 649. Subtract this from 901 to get 901 - 649 = 252.
Note that it does not matter if we include "100" or "1000" in our set since we'll be subtracting them anyway (instead of 901 - 649 we'll get 900 - 648 or 899 - 647 = 252).
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