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In a prize drawing at a fundraiser, you choose three different numbers from 1 to 8. How many ways are there to choose three numbers?
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This is a combination problem :) First, you must understand the concept of "factorial" It's denoted by a "!" n! = n x (n-1) ...... x 1 For example 4! = 4 x 3 x 2 x 1 = 24 Note that 1! = 1 = 0! All set? Good. Now a combination, denoted by nCr means how many ways you can select r items from a set of n items. Its formula is given by: \[nCr = \frac{n!}{(n-r)!r!}\]
Now you have 8 numbers and are choosing 3 So you have to figure out 8C3 Good luck! And feel free to ask me if you get stuck :)
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