# Math Help - Counting products of a set of numbers

1. ## Counting products of a set of numbers

How many different numbers can be made as the product of two or more of the numbers {3,4,4,5,5,6,7,7,7}?

If all the numbers in the set were prime, every product would be unique and I wouldn't have to worry about overlaps.

I am not sure how to account for the possible repeats?

How would I begin a problem like this? Any hints please?

2. ## Re: Counting products of a set of numbers

Originally Posted by Jame
How many different numbers can be made as the product of two or more of the numbers {3,4,4,5,5,6,7,7,7}?
If all the numbers in the set were prime, every product would be unique and I wouldn't have to worry about overlaps.
Consider the product $3^a\cdot 4^b\cdot 5^c\cdot 6^d\cdot 7^e$.
Now you want $0\le a\le 1,~0\le b\le 2,~0\le c\le 2,~0\le d\le 1,~\&~0\le e\le 3$ while $a+b+c+d+e\ge 2.$

3. ## Re: Counting products of a set of numbers

Thanks for the very clear explanation! I have a very similar problem, but it deals with addition and I'm not sure how to account for this.

How many different positive integers can be obtained as a sum of two or more of the numbers

{1,3,5,10,20,50,82}

I don't have the primes to help me here.

Its not (7 choose 2) + (7 choose 3) + ... etc

I can already think of 1 repeat.

83 = 82 + 1 = 10+20+50+3

4. ## Re: Counting products of a set of numbers

Originally Posted by Jame
Thanks for the very clear explanation! I have a very similar problem, but it deals with addition and I'm not sure how to account for this.
How many different positive integers can be obtained as a sum of two or more of the numbers
{1,3,5,10,20,50,82}
I don't have the primes to help me here.
Its not (7 choose 2) + (7 choose 3) + ... etc
I can already think of 1 repeat.
83 = 82 + 1 = 10+20+50+3
You and I must be reading this question differently.
As I read it we could have $7^2$ but we could not have $4^1$.
As I count it you have $2\cdot 3\cdot 3\cdot 2\cdot 4-6=138$ possible products.