# Thread: Basic Counting Methods

1. ## Basic Counting Methods

The question is
How many integers are between 0 and 40? my answer 41
How many of these integers are divisible by 2? my anser 20

How many unordered pairs of these integers are whose difference is 5?

Here I do not know where to start? An unordered pair is {#1, #2}, so I think I have two sets of numbers
set one are all numbers possible for #1 and
set two are all numbers possible for #2

First I am not sure how to treat the sets, because the pairs are unordered. So, What I understand is that in the first place could be a 30 and in the second a 5 or the other way around.

Then, I do not know how to express the "difference is 5"

Thanks for any help.

2. Originally Posted by DBA
The question is
How many integers are between 0 and 40? my answer 41
How many of these integers are divisible by 2? my anser 20
Has either your instructor or your textbook discussed the meaning of between?
The way either uses that word will change the answer to the first question.
For example, I do not consider 0 to be between 0 & 40.
It is clear that 5 is between 0 & 40, but 40 is not.
However, I have seen people who do disagree with me.
I know in test prep writing one uses the pharse ‘from 0 to 40’ if we intend inclusion.

3. Between to numbers include the numbers at the edges. So 0 to 40, includes the 0 and the 40 in my textbook.

4. Originally Posted by DBA
Between to numbers include the numbers at the edges. So 0 to 40, includes the 0 and the 40 in my textbook.
O.K. So we have a list of integers from 0 to 40.

Originally Posted by DBA
The question is
How many unordered pairs of these integers are whose difference is 5
The 'smallest' unordered pair is $\{0,5\}$.
The largest is $\{35,40\}$.
Another such pair is $\{8,13\}$.
So how many altogether?

5. So for my first number of the pair I have numbers from 0 to 35, which are 36 possibilities and for the second number of the pair I have numbers from 5 to 40, which are 36 possibilities.
My total possibilities to combine them would be 36*36. is that correct?

Would that include that I can have a pair like {13,8}? Since it is unordered the order should not matter, right?
Thanks

6. Originally Posted by DBA
So for my first number of the pair I have numbers from 0 to 35, which are 36 possibilities and for the second number of the pair I have numbers from 5 to 40, which are 36 possibilities.
My total possibilities to combine them would be 36*36. is that correct?
Would that include that I can have a pair like {13,8}? Since it is unordered the order should not matter, right?
Why in the world would you square the number?
An unordered pair is simply a set of two terms.
So $\{13,8\}=\{8,13\}$, one unordered pair.

7. mmh a wrong thought I guess. I wrote them out and now it is clear that there are 36 combinations possible. Thanks for your help!