Calculate (without bruce force) probability that Diana leaves at the same time as AT LEAST one of her brothers? 
Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that Diana leaves at the same time as AT LEAST one of her brothers?

Let there be groups $A,B,C$.
Let there be three positions, and each position is either one or zero.
If it's one they're in some fixed group. If it's zero, they're absent from that fixed group.
000 means they're all not in some fixed group.
Let Diana be in the first position, Bruce the second, and Clerk the third.
The favorable cases regardless of the groups are 110,101, and 111.
There are 4 other cases.
001
010
000
100
I can list everything out in this way but it's too brute force.
How do you calculate the probability without counting the discrete cases?

 A: Assuming the three siblings are randomly distributed to the three groups independently with equal probabilities, we can count the number of invalid cases and subtract from the total number of cases to find the number of valid cases.
The total number of cases is:
$3$ positions for Diana $*$ $3$ positions for Bruce $*$ $3$ positions for Clerk $= 27$ total cases.
The number of invalid cases (where Diana is not with either brother) is:
$3$ positions for Diana $*$ $2$ remaining positions for Bruce (as he cannot be in the same group as Diana) $*$ $2$ remaining positions for Clerk $= 12$ total cases.
Thus the answer is $\frac{27-12}{27} = \frac{5}{9}$

Alternatively we can remove that assumption for fun. Suppose the three groups are of sizes $a, b,$ and $c$, and Diana, Bruce, and Clerk are just three students of a total $a + b + c$ students. If the groups are randomly created, what's the probability that Diana is in the same group as at least one of her brothers?
We can conceptualize the problem as follows:
Place all $a+b+c$ students in a line in a random order.
Then take the first $a$ students as the first group, the next $b$ students as the second group, and the final $c$ students as the third group.
Going back to what we did before, the total number of ways to place Diana, Bruce, and Clerk in the line is $(a+b+c)(a+b+c-1)(a+b+c-2)$.
The number of invalid cases is:
$a(b+c)(b+c-1)$ for the case that Diana is in the first group,
$b(a+c)(a+c-1)$ for the case that Diana is in the second group, and
$c(a+b)(a+b-1)$ for the case that Diana is in the third group.
So the final probability that Diana is in a group with at least one of her brothers is
$$1 - \frac{a(b+c)(b+c-1) + b(a+c)(a+c-1) + c(a+b)(a+b-1)}{(a+b+c)(a+b+c-1)(a+b+c-2)}$$
