The house numbers on a street each have 4 digits that range from 1000 to 9999. How many of these house numbers have exactly 2 digits that are 5?

Question

My problem with this question is the fact that the first number doesn't have as many choices as the last 3. I'm sure it's easier than I thing it is but I just can't get around it. I have 4C2 ways of picking where to put my 5s, but I now don't know how many ways I can pick and order the last 2 numbers because I won't know whether or not one of my 5s is in the first spot.

Answer 1

If the first digit is 5:

1 combination for the first place. ->1

2 numbers other than five need to be picked with repetition. ->$9^2$

3 positions are available between these 2 numbers to place the remaining "5" (._ . _ .)$ ->3

So $9^2 \times 3$ combinations here.

If the first digit isn't 5:

8 numbers available for the first digit -> 8

2 spots out of 3 for the two "5" digits (order irrelevant) -> 3

9 numbers available for for the remaining spot -> $9$

So $8 \times 3 \times 9 $ combinations here.