zepdrix another question: nCr=n!/(n-r)!*r! I don't understand why we add the r! in the denominator , I mean for example if we have 53 cars racing, and we want to know the possible combinations of the three racers that arrive first, we'd do : 53!/(53-3)! , we divide by 50! cuz we only need 3 places right, so all we'll be getting in the end is : 53*52*51 , so why the 3! ?

Question

zepdrix another question: nCr=n!/(n-r)!*r! I don't understand why we add the r! in the denominator , I mean for example if we have 53 cars racing, and we want to know the possible combinations of the three racers that arrive first, we'd do : 53!/(53-3)!  , we divide by 50! cuz we only need 3 places right, so all we'll be getting in the end is : 53*52*51 , so why the 3!  ?

etherealsarah · Answer

@S4Sensitiveandshy can u help please! :)

solomonzelman · Answer

Because in combinations the order doesn't matter. (So, «a, b, c, d» and «b, d, c, a» for example, are the same combinations).

So, we divide by the factorial of r!
(r is the number that we choose out of n).

solomonzelman · Answer

In permutations, where order does matter, we would not divide by r!, rather,
nPr = n!/(n-r)!

etherealsarah · Answer

I can't understand how the "importance" of the order is eliminated when we divide by r! ?

etherealsarah · Answer

@SolomonZelman