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Math Help - Counting products of a set of numbers

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    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?
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    Re: Counting products of a set of numbers

    Quote Originally Posted by Jame View Post
    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.
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    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
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    Re: Counting products of a set of numbers

    Quote Originally Posted by Jame View Post
    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.
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