+0  
 
0
845
1
avatar

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

 Jul 14, 2017

Best Answer 

 #1
avatar+23246 
+1

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

C(5, 2)  =  5! / ( 2! x 3! )  =  10

 Jul 14, 2017
 #1
avatar+23246 
+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

5 Online Users

avatar
avatar
avatar
avatar