Probability of selecting 3 students being in different classes I have come across this problem and I can't quite figure out the solution.
A class of 60 students is divided into 3 groups of 20 randomly. Three friends are in this year group. Find the probability that all three of them are in different classes.
The answer given is 800/3422
However I work out that:
A) The probability that all of them are in the same class is: 20/60x19/59/18/58x3x3 where the first three is for all the classes and the second three is to account for the 3 different students. This is 1026/3422
B) I find the probability that 2 are on the same class and one is on a different class. 20/60x19/59x40/58x6. 6 is for all the possible combinations of classes. I get 1520/3422
C) Last step is that all the cases should add up to 1 so I get that the probability that the students are in different classes is 876/3422.
I obviously have made a mistake. Can anyone spot it? Thank you
 A: This can be calculated more simply as:
$$P(Friend1 \in any\ class) \times P(Friend2 \notin class_{friend1}) \times P(Friend3 \notin class_{friend1} \& Friend3 \notin class_{friend2})$$
$P(Friend1 \in any\ class) = 1$
$P(Friend2 \notin class_{friend1}) = 40/59$ (40 out of the remaining 59 places meet this criterion)
$P(Friend3 \notin class_{friend1} \& Friend3 \notin class_{friend2}) = 20/58$ (20 out of the remaining 58 places meet this criterion)
$=> P(Friends\ are \ in\ 3\ different\ classes) = 40/59 \times 20/58 = 800 / 3422$
A: To do this in a straight forward way, friend1 can land in 1 of the 60 positions, 20 of which is in class1; then, friend2 can land in 1 of the remaining 59 positions, 20 of which is in class2; similarly, 20/58 positions for friend3 in class3.
So probability of (friend1 in class1) & (friend2 in class2) & (friend3 in class3) is (20/60)x(20/59)x(20/58)=800/20532.
Now consider
(friend1 in class1) & (friend3 in class2) & (friend2 in class3)
(friend2 in class1) & (friend1 in class2) & (friend3 in class3)
(friend2 in class1) & (friend3 in class2) & (friend1 in class3)
...
etc

You need to multiply by 6 possible positions. Which gives you (800/20532)*6=800/3422

I think you got it wrong in 2 places. In A there is 1 possible combination of 3 friends, so only multiply 3 once for classes. In B the combination of classes is 3 (not 6 since the 2 possible remaining classes for 3rd friend is covered in the 40) and you forget the combination-of-2-friends which is 3.
so probability for 3 friends in the same class is (20/60)x(19/59)x(18/58)x3, 2 friends in the same class is (20/60)x(19/59)x(40/58)x3x3. The result is the same.
