# Number of combinations

#### aron101782

##### New member
If a sequence of numbers is as follows
A B C
such that
0<A<2
0<B<2
0<C<1

What is a convenient way to determine the number of combinations in this sequence

#### JeffM

##### Elite Member
What kind of numbers?

Please give the exact and complete language of the problem you are trying to solve. What you have written makes no sense.

#### Dr.Peterson

##### Elite Member
I'm not quite sure what you are asking. As I read it, you want to know the number of different sequences A, B, C such that 0<A<2, 0<B<2, and 0<C<1. But if A, B, and C are required to be integers, then the answer is "none" (there is no integer C between 0 and 1); and if they can be any real numbers, then the answer is "infinitely many", which is not very interesting.

Also, you can't be using the word "combinations" in its technical sense, as clearly order matters.

Can you clarify the question?

#### Jomo

##### Elite Member
If a sequence of numbers is as follows
A B C
such that
0<A<2
0<B<2
0<C<1

What is a convenient way to determine the number of combinations in this sequence
Since A is between 0 and 2 there are infinite number of numbers that A can be
Since B is between 0 and 2 there are infinite number of numbers that B can be
Since C is between 0 and 1 there are infinite number of numbers that C can be

Continue from here

#### aron101782

##### New member
I'm sorry I meant
0<=A<=2 : 0 1 2
0<=B<=2 : 0 1 2
0<=C<=1 : 0 1

What is a convenient way to determine the number of possibilities for ABC

#### Dr.Peterson

##### Elite Member
That makes a huge difference! This is one of many reasons we ask to see your own thoughts on a problem (here). It helps us know what you mean.

How many ways are there to choose A?

For each of those, how many ways are there to choose B?

How many ways, therefore, are there to choose a pair A, B?

Now continue.

#### pka

##### Elite Member
I'm sorry I meant
0<=A<=2 : 0 1 2
0<=B<=2 : 0 1 2
0<=C<=1 : 0 1

What is a convenient way to determine the number of possibilities for ABC
The terms $$\displaystyle ABC$$ count is $$\displaystyle 3\cdot 3\cdot 2=18$$
I not know what more you are asking.

#### aron101782

##### New member
so there would be 18 combinations for ABC

#### JeffM

##### Elite Member
Yes. The fast way to figure it out is pka's way, but you can also do it by systematic listing.

1 000
2 001
3 010
4 011
5 020
6 021
7 100
8 101
9 110
10 111
11 120
12 121
13 200
14 201
15 210
16 211
17 220
18 221