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Is there a special way for working out these types of questions?

Guest Jul 14, 2017

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 #1
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This is a "combinations" problem  --  choosing an unordered group of 2 from a group of 5.

C(5, 2)  =  5! / ( 2! x 3! )  =  10

geno3141  Jul 14, 2017
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 #1
avatar+17652 
+1
Best Answer

This is a "combinations" problem  --  choosing an unordered group of 2 from a group of 5.

C(5, 2)  =  5! / ( 2! x 3! )  =  10

geno3141  Jul 14, 2017

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