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# Hi, do you guys mind helping me? TY!

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Ms. Q wrote the numbers 1 - 14 inclusive. She chose 13 random numbers and multiplied them. How many possible products could she have gotten?

I tried this problem, and I got 182 by multiplying 13 and 14, but I am not really sure. Do you guys mind helping me?

May 24, 2023
edited by Guest  May 24, 2023

#1
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here are 14×13×12×11×10×9×8×7×6×5×4×3×2=343,248,960 total ways to choose 13 numbers from 1 to 14. However, some of these products will be the same. For example, if she chooses 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13, then the product will be the same as if she chooses 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1.

There are 13! ways to order 13 numbers. However, half of these orders will result in the same product. Therefore, there are 13!/2=28,657,600 unique products.

May 24, 2023
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14 C 13 = 14 possible products [Don't overthink it !]

May 24, 2023
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um, whose right? Sorry, I either overthought it, or underthought it, and I'm not sure which. Thanks for answering though!

May 24, 2023
#4
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Take a smaller example so that you can understand it:

Take 5 numbers: 1, 2, 3, 4, 5

5 choose 4 ==5 combinations. These 5 combinations are as follows:

(1234, 1235, 1245, 1345, 2345) >>Total = 5 combinations.

The product of each combination is EXACTLY the same, no matter how you re-arrange them:

Take the first one: 1 x 2 x 3 x 4 ==24. Or you can have: 4 x 1 x 3 x 2 ==24. Or you can have: 3 x 1 x 2 x 4 ==24......and so on. You can arrange this one in: 4! ==24 different arrangements or permutations. But all 24 will give you exactly the same product, which happens to be 24. The next one: 1235 ==1 x 2 x 3 x 5==30. Again, you can have 4! ==24 different arrangement, but will always have the same product of 30.

So, in conclusion you will have 5 products as follows: 24 , 30, 40, 60, 120, despite the fact that you could re-arrange them in: 4! x 5 ==120 re-arrangements or permutations. But all those permutations will have only 5 products of: 24, 30, 40, 60, 120 even though you will 24 copies of each. Or 24 permutations of 24, 24 permutaions of 30, 24 permutations of 40, 24 permutations of 60 and 24 permutations of 120.

Guest May 24, 2023
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There are 14 choices for the first number, 13 choices for the second number, and so on. This gives us a total of 141312111098765432*1 = 387,420,489 possible products.

However, some of these products will be the same. For example, if Ms. Q chooses the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13, the product will be 14!, which is 87178291200. If she chooses the numbers 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14, the product will also be 14!.

There are 14 ways to choose 13 numbers from 1 to 14, so there are 14 copies of each product. This means that we need to divide the number of possible products by 14.

This gives us a total of 387,420,489 / 14 = 27,671,464 possible products.