Question 1024915: I have four numbers: 2, 92, 79, and 71. Without using any number twice, and always using the four numbers, I need to come up with a list of every possible combination. I know that is going to add up to a very large amount of combinations, but so far, every other place I checked failed miserably. I guess it's up to you to save the day now. Good luck, and thank you!
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
I have four numbers: 2, 92, 79, and 71. Without using any number twice, and always using the four numbers, I need to come up with a list of every possible combination. I know that is going to add up to a very large amount of combinations, but so far, every other place I checked failed miserably. I guess it's up to you to save the day now. Good luck, and thank you!
Solution:
The number of combinations for taking any number of object (without repetition) from a pool of n is  ( equal to the size of the power set).
It seems from the context that at least one must be used, which reduces the number of combination by 1.
So the number of combinations is  .
However, if you say there is a very large number of combinations, then you might mean doing operations between the numbers. In that case, more rules are required, for example: the permissible operators (+,-,*,/ or more), the presence of parentheses, and how many levels of nesting parentheses, etc.
If these rules are provided, please make a new post so everyone can have a chance to help you.
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