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- For example, if you have z people and know that choosing two of them would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would yield exactly 15 different groups of two.
Select 2 people from z people. We have \(C^2_z\) different ways.
\(C^2_z=\frac{z!}{2!(z-2)!}=\frac{z(z-1)}{2}=15 \implies z(z-1)=15 \implies z=6\)
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- in this case, x must equal 6, because the only way to obtain 56 groups if choosing 3 is to choose from a group of 8. Since the statement
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