Question

How many four digits number are there whose decimal notation contains not more than two distinct digits ?

Answer 1

I think the easiest way to approach this question, is to find how many 4 digit numbers there are, and subtract however many have completely different digits, and how many have 3 different digits. Unfortunately, there's still going to be a lot of case work (assuming that zero is still not allowed for the first digit).

Answer 2

First, how many 4 digit numbers are there total?

Answer 3

9999-1000=8999...

Answer 4

There should be 9000. 9 options for first digit, \(10^3\) for the remaining three digits. Now we need to figure out how many don't have any repeated digits.

Answer 5

First, 9 options for the first digit. Now, notice that there are still 9 options for the second digit since we can also use 0 now. Then there are 8 possibilities, and 7 for the units digit. So there are\[9\cdot9\cdot8\cdot7\]possibilities where all the digits are distinct.

Making sense so far?

Answer 6

didn't get why the 2nd is again a 9....

Answer 7

The first digit could be chosen from the set \(\{1,2,...,9\}\). Let's suppose that we chose \(i\). Then the second digit is chosen from the set \(\{0,1,2,...,9\}-\{i\}\). So everything in the set \(\{0,1,2,...,9\}\), except the number \(i\). There are 10 digits in the original set, and we subtract 1. That leaves us with 9 digits to choose from.

Answer 8

understood it... :)

Answer 9

Thank you :)

Answer 10

Great. Now we're left with the case where two digits are the same, but the rest are distinct. This case we'll have to break up into 2 more case.

Case 1: The repeated number is used for the first digit. There are 9 options for the first digit. Now we have to choose which of the remaining 3 places are the same. So that's another 3 options. Whatever place we choose for this though, it only has a single possibility for what number it is (the same as the first digit).

Now we have two remaining digits. Each of these must be different from the other digits. But since it was the very first digit that was repeated, we can still choose 0. This results in 9 possibilities for one digit, and 8 for the final digit.

So in this case, there are\[9\cdot3\cdot9\cdot8\]possibilities.

Answer 11

Case 2: The first digit is not repeated. Then there are still 9 options for the first digit. We're left with three places, 2 of which repeat. So there are \(\binom{3}{2}=3\) ways to choose those two places. Since we can include 0 again, there are 9 possibilities for what number they are.

Finally, we have one digit left, and it has 8 possibilities. So in this case there are\[9\cdot3\cdot9\cdot8\]possibilities again.

Is there anything I should explain in greater detail?

Answer 12

the answer provided is 576.. Will this end up to 576...

Answer 13

Let's see. The last step is to subtract the number of possibilities where digits don't repeat, or where one is repeated from all possibilities. This is equal to\[9000-9\cdot9\cdot8\cdot7-2\cdot9\cdot3\cdot9\cdot8\]and if you put this in a calculator, it is indeed 576.

http://www.wolframalpha.com/input/?i=9000-9%C3%979%C3%978%C3%977-2%C3%979%C3%973%C3%979%C3%978

Answer 14

Got it.. thank you very much...:)

Answer 15

You're welcome. Remember that you will often have to resort to case work as above. Just make sure everything is making sense, and you'll be fine. No matter what method you use, it should all come out to be the same thing at the end.