A tough math problem?

Question

How many four digit numbers contain the digit pattern 85 at least once and only once?

Answer 1

Let first two digits be 85, then next two digits can be 00 to 99 minus 85 thus totalling 99 ways.
Let the 2nd and 3rd digits form 85, then there are 90 ways in all in which the first digit is any from 1 to 9 and the last digit is any from 0 to 9.
Let the last two digits be 85. Then there are in all 89 ways with the first digit being any from 1 to 9, and the second being any from 0 to 9 and the first two are not 85.
Total = 99 + 90 + 89 = 278.
I have edited my answer after seeing the answer of snogledk.

Answer 2

two different questions here.

Just work through them. The interesting ones are all numbers on the forms 85XX, X85X and XX85. In the first two forms, X can be any digit from 0 to 9, giving us a total of 100 possibilities each. In the third form, the first X cannot be 0. So, there are only 90 possibilities here. All together: 290 numbers.

However, we have counted the single number 8585 twice, so the real solution is 289 numbers.

Answer 3

85xx, x85x, xx85
In the 1st case the 1st x cannot be an 8 followed by a 5, so there are just 99 possibilities
In the 2nd case,each can be any of 10 digits so total is100
3rd case is same as 1st case so 99possibilities
Total = 99+99+100 = 298