Probability. This is the limit. Why is my answer not correct again???????It's really simple. Please see.

A bag contains 20 chocolates, 15 toffees, and 12 peppermints. If three sweets are chosen at random, what is the probability that they are all different.

Ain't it obvious that only 12 combinations could be made if they were all to be different? Cux there are only 12 peppermints? So i divided 12 by all possible combinations ie 47C3. But it's wrong. How??? Am I never gonna get the hang of it?!!!!

Please give detail. thanks.

Number of ways of selecting r number of items out of n is given by nCr = n! / r!*(n-r)!
1 chocolate can be selected in 20C1 ways
1 toffee can be slected in 15C1 ways
1 pepperment can be selected in 12C1 ways
3 sweets can be selected in 47C3 ways
Required probability = P(choosing different sweets) = 20C1*15C1*12C1 / 47C3
= 3600 / 16215
= 0.222

Sorry did it wrong, edit:

The total number of outcomes possible = 47C3
= 47*46*45 / 6
= 16215 possibilities.

All different pieces would be = 20C1*15C1*12C1
= 20*15*12 = 3600

Therefore the probability of selecting 3 different pieces is: 3600/16215 = 240/1081

Jeris G has it now