I seem to be missing something
- Thread starter Loki123
- Start date
If the question is how many integers > 999 and < 10000 can be formed by using just two of the ten digits, then think like this.
We have three basic cases.
All four digits are alike.
Three digits are alike.
There is a pair of digits, with each of the pair being used twice.
Within those cases, we must address the issue that the thousands digit must not be 0.
We have three basic cases.
All four digits are alike.
Three digits are alike.
There is a pair of digits, with each of the pair being used twice.
Within those cases, we must address the issue that the thousands digit must not be 0.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,749
You appear to be trying to count the number of 4-digit numbers that consist of no more than two distinct digits (and do not start with 0).
It's easy to count those using only one digit. (Don't include 0000.)
Then you need to choose two digits (10C2), and count the ways to use them (2^4), but exclude those which use only one of the two digits, which are already counted.
Finally, you need to subtract all the numbers that have been counted, but which start with 0. These would be the arrangements of 3 digits, one of which is 0. (There may be an additional adjustment.)
I think one error you made is in combining the 45 and the 10, which behave differently.
When you say your answer is incorrect, is that because you have been told a "correct" answer, or for some other reason?
It's easy to count those using only one digit. (Don't include 0000.)
Then you need to choose two digits (10C2), and count the ways to use them (2^4), but exclude those which use only one of the two digits, which are already counted.
Finally, you need to subtract all the numbers that have been counted, but which start with 0. These would be the arrangements of 3 digits, one of which is 0. (There may be an additional adjustment.)
I think one error you made is in combining the 45 and the 10, which behave differently.
When you say your answer is incorrect, is that because you have been told a "correct" answer, or for some other reason?
i have suggested answers, mine is not one of them
Still incorrect, but I really don't know why now.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,749
Please start telling us the entire problem, rather than giving us only what you think we need. If you are saying there is a list of choices, that is part of the problem. (Even a suggested "correct" answer is part of your problem, because it tells us what you are comparing to, and it could be wrong.)
Do you not notice that your 45 includes cases where all are one digit or the other? You need to exclude those. Think similarly about the 8; there are several errors there (such as that 0aab has three digits, not 2!).
Combinatorics requires constant thinking and rethinking. This is why I mentioned adjustments.
What is the problem asking?
If it is asking for the number of ways to form an integer in the thousands using exactly two distinct digits, my answer is 243. Where does that come from?
If it is asking for the number of ways to form an integer in the thousands using at most two digits, my answer is 576. Where does that come from?
As Dr. Peterson says, combinatorics requires careful thinking before applying formulas. Why in the world are you using combinations when order matters: isn’t 1122 a different number than 2211?
If it is asking for the number of ways to form an integer in the thousands using exactly two distinct digits, my answer is 243. Where does that come from?
If it is asking for the number of ways to form an integer in the thousands using at most two digits, my answer is 576. Where does that come from?
As Dr. Peterson says, combinatorics requires careful thinking before applying formulas. Why in the world are you using combinations when order matters: isn’t 1122 a different number than 2211?
i am not using combinations for numbers themselves. i am using combinations to pick pairs of numbers as the order doesn't matter there. then i am using variations to see how many different numbers i could get using said pairs.
But order does matter.
Consider the question of how many numbers in the thousands can be made using exactly two digits.
There are exactly 9 ways to choose the thousandth digit because 0 must be excluded.
There are exactly 9 ways to choose the digit that differs from the thousands digit. We are dealing with 81 possible ordered pairs rather than the 45 unordered pairs you came up with.
So why do I say we get 243 different numbers? Because we can choose to repeat the digit used in the thousands position in the hundreds position or the tens position or the units position. 3 * 81 = 243.
As Dr. Peterson said, combinatorics requires thinking carefully about precisely specified problems. Of course, this problem as you have stated it is not precisely specified, which is why I cam up with two different answers. Why don’t you try to replicate the reasoning that gets to an answer of 576? I have now given you two posts that contain clues.
Consider the question of how many numbers in the thousands can be made using exactly two digits.
There are exactly 9 ways to choose the thousandth digit because 0 must be excluded.
There are exactly 9 ways to choose the digit that differs from the thousands digit. We are dealing with 81 possible ordered pairs rather than the 45 unordered pairs you came up with.
So why do I say we get 243 different numbers? Because we can choose to repeat the digit used in the thousands position in the hundreds position or the tens position or the units position. 3 * 81 = 243.
As Dr. Peterson said, combinatorics requires thinking carefully about precisely specified problems. Of course, this problem as you have stated it is not precisely specified, which is why I cam up with two different answers. Why don’t you try to replicate the reasoning that gets to an answer of 576? I have now given you two posts that contain clues.
i am confused.
01
02 12 23 34 45 56 67 78 89
03 13 24 35 46 57 68 79
04 14 25 36 47 58 69
05 15 26 37 48 59
06 16 27 38 49
07 17 28 39
08 18 29
09 19
see 45 pairs.
if i pick 1 and 2, it's like i picked 2 and 1. order doesn't matter in the sense what numbers i pick, where i put them is what matters.
this makes sense but wouldn't we then be picking two of the same. I even counted aside the 45 possibilities
01
02 12 23 34 45 56 67 78 89
03 13 24 35 46 57 68 79
04 14 25 36 47 58 69
05 15 26 37 48 59
06 16 27 38 49
07 17 28 39
08 18 29
09 19
see 45 pairs.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,749
I don't see that you have taken my suggestions into account. How are you avoiding counting the same number more than once?
You may want to write out the meaning of each case you calculate in words, rather than using a brief notation that is ambiguous, so neither we nor you can be sure exactly what you are including.
Combinations are appropriate as part of the work, even in problems where order will matter. That's not the problem here; it's not being careful about exactly what is being counted.
My method (which gives the answer 576) is similar to Loki's, but has a couple significant adjustments. I counted numbers with one distinct digit (not 0), and those with two distinct digits, including 0, and then subtracted those with two distinct digits, one of which is 0, with 0 in the first digit. I do not get 720 or 72.
But there are probably many other ways to do it!
i really don't get your method. can you give me some more clues? this is literally the best I can do at the moment.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,749
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.
so?
it says at most two digits
so i can use only one?
this doesn't really make sense to me. how can you just subtract from variations?
I fully agree that [math]\dbinom{10}{2} = 45.[/math] Where I do not agree is that starting with the number of ways to select two distinct items from ten distinct items without repetition and without regard to order when the problem involves selecting four digits where repetition is required and where order does matter leads to an intuitively obvious solution.
Dr. Peterson and I agree that frequently there are multiple methods to solve a given problem. We also agree that problems in combinatorics require careful thought. And I cannot dispute that what is most intuitive to me may not be intuitive to others.
Our first question here is what does the problem mean.
Based on the answers given, I think the problem is asking "How many integers are there that are in the thousands and that are formed by at most two of the ten digits." But to demonstrate an intuitive way to attack the problem, I selected the simpler question of "How many integers are there that are in the thousands and that are formed by exactly two of the ten digits." Notice that the latter question helps answer the former question.
How many ways can we select the digit in the thousand's position? [imath]\dbinom{10}{1} = 9.[/imath]
How many ways can we select the other of the two digits? [imath]\dbinom{10}{1} = 9.[/imath]
How many places can we repeat the digit in the thousand's position? [imath]\dbinom{3}{1} = 3.[/imath]
How many integers subject to the constraints? [imath]9 * 9 * 3 = 243.[/imath]
i get it, but this isn't a suggested answer #9
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?