How many different ways can Joe eat the candy

Question

Joe is eating colored candy and he has 5 orange candies and 5 yellow candies left. how many different ways (color patterns) can joe eat the candy?

Answer 1

If you imagine Joe lining up the candies and numbering them 1 thru 10 before he eats them in that order, then the number of color patterns is uniquely specified by the number of different placements of orange candies.  That is, once Joe decides which positions he's going to put the 5 orange candies in, then the positions of the 5 yellow candies are fixed.  So how many choices does Joe have to place the orange candies?  In essence, Joe is choosing 5 locations for the orange candies out of 10 available locations.  So the number of choices is 10 choose 5, which is written ₁₀C₅ and  

 

₁₀C₅ = 10!/(5!(10-5)!) = 10*9*8*7*6/(5*4*3*2) = 1*3*2*7*6 = 42*6 = 252  

 

Note that there is some symmetry in this problem, meaning that you could have solved for the number of placements of yellow candies instead and gotten the same answer (obviously, since ₁₀C₅ = ₁₀C₅).  But the problem is symmetric even if the number of orange candies is different than the number of yellow candies.  So if there were 3 orange candies and 7 yellow candies, you could get the same answer by computing ₁₀C₇ or ₁₀C₃, and this is the reason that the choose function is symmetric about n/2.