Question 838133: how many 3 digit integers greater than 499 have exactly two repeated digits
Found 2 solutions by LinnW, richard1234:
Answer by LinnW(1048)   (Show Source):
You can put this solution on YOUR website! There are ten numbers with at least 2 repeated
digits between 500 and 599,
500,511,522,533,544,555,566,577,588,599.
Similarly there are 10 numbers
between 600 and 699 , 700 and 799, 800 and 899 and 900 and 999
This gives us 50 numbers.
We need to subtract 5 from 50 since
555,666,777,888 and 999 have three repeated digits.
So we end up with 45 numbers meeting the criteria.
We also need to account for numbers like,
550,551,552,553,554,555,556,557,558,559
There are 10 numbers above but 555 must be excluded.
So we are left with 9 numbers of the form 55n .
There are also 9 numbers of the form 66n,77n,88n,99n
We have 45 numbers of the forms 55n,66n,77n,88n,99n
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So the overall count is 45 + 45 = 90
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! For the 500s there are 500, 511, ..., 599 as well as 550, 551, 552, ..., 559 giving us 10+10 = 20 numbers. However we are counting 555 twice, but we want to count it zero times, so we subtract 555 twice, giving us 18 numbers between 500 and 599 (inclusive) with two repeated digits.
Same thing holds for 600-699, ..., 900-999, giving us 18*5 = 90 numbers.
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