I seem to be missing something

[Loki123]

Oh for goodness sake, I SPECIFICALLY said that, based on the suggested answers given, I do not believe that what I answered in the preceding post was the intended question. What I gave you you in my immediately preceding post was a method that you could use to attack the more general question that I believe was intended. I also said that the answer given in my immediately preceding post assists in answering the broader question.

Do you even bother to read the posts made by the helpers?

Sorry, it was late I must have overseen it.

 

[Loki123]

To count the number of four-digit numbers using exactly two digits, and allowing the first digit to be zero, you have C(10,2) = 45 ways to choose the two digits, A and B. Then there are 2^4 ways to choose one of those two digits for each of the 4 places. But that includes the numbers AAAA and BBBB, which use only one digit, not two. So you have to subtract those; rather than 16 ways to use the two digits, there are 2^4-2 = 14. So multiply 45 by 1, not 16.

This is not that hard to figure out! You just have to be willing to think about the implications of every step you take.

Where am I going wrong

 

[JeffM]

If you start from your [imath]\dbinom{10}{2}=45[/imath] and then multiply by [imath]2^4=16[/imath], you get 720 possibilities. But that is far too large because it includes numbers like 0000, 0707, 0007, 0777, etc. How many of those are there?

 

[Loki123]

If you start from your [imath]\dbinom{10}{2}=45[/imath] and then multiply by [imath]2^4=16[/imath], you get 720 possibilities. But that is far too large because it includes numbers like 0000, 0707, 0007, 0777, etc. How many of those are there?

72

 

[Dr.Peterson]

I give up.

Where am I going wrong View attachment 32605

That could be another valid way to do it; but how are there 90 ways to make "aaaa"? You must mean something different than I would mean by that. Please state in words what each case you count means, rather than expecting us to figure out your notation.

Also, I'm not sure if you're taking into account the fact that you're counting some numbers multiple times (because they can be formed by taking all the same digit out of different pairs -- which is why I proposed counting numbers using one digit separately).

 

[JeffM]

I want to acknowledge that Dr. Peterson and I are thinking along different lines. That is no problem; different people think in different ways, but it may make it hard for you to follow.

It is definitely possible to start by calculating a number that is obviously too big and then subtracting adjustments, taking care that no adjustments involve double counting.

I, however, like to think in terms of adding the numbers of mutually exclusive sets rather than calculating a set that we know to be too large and then trying to make adjustments. That is just my preferred way of thinking. It is not the exclusive way to think.

How many ways can you select four digits that are all alike but excluding the digit 0? Example is 8888.

How many ways can you select an initial digit that is not zero and then a different digit that will be used three times? Example is 2333.

How many ways can you select an initial digit that is not zero, use it two more times, and use a different digit once? Examples are 2232 and 7077.

How many ways can you select an initial digit that is not zero, use it one more time, and use a different digit twice. Examples are 2020, 7755, 9119.

Does any of that involve double counting?

Are those all the different ways to denote an integer in the thousands using at most two decimal digits?

If so, our answer is the sum of of the numbers in those mutually exclusive but exhaustive sets.

You can do it your way, but then you have to make adjustments for any results that do not fit the requirements and all instances of double counting. I tend to make too many logical errors when adjusting something known to be wrong at the start (of course my way is not guaranteed to avoid logical errors).