Multinomial theorem with imposed conditions 
The number of ways in which 12 identical balls can be grouped in four marked non-empty sets $P, Q, R, S$ such that $n(P) < n(Q)$ is?

The answer to the above problem is the number of positive integral solutions of the expression $$P + Q + R + S = 12$$
when $P<Q$. I know that with no conditions imposed, the number of solutions is $^{12-1}C_{4-1} = ^{11}C_3$. How would one account for the condition when $P<Q$? Please explain the concept in detail in addition to providing a solution to the problem (which is given just as an example).

So first of all, the solution without that condition is $\binom{12-1}{4-1}$. Since the sets are all non - empty.
Notice that $Q-P$ is a positive integer as well. The number of solutions with condition imposed is the number of positive integer solutions of
$2P+(Q-P)+R+S=12$
Since the number is small, You can count solutions for $2P=2,4,6,8$.
$2P=2,\ \ (Q-P)+R+S=10$ there are $\binom{10-1}{3-1}=36$ solutions
$2P=4,\ \ (Q-P)+R+S=8$ there are $\binom{8-1}{3-1}=21$ solutions
$2P=6,\ \ (Q-P)+R+S=6$ there are $\binom{6-1}{3-1}=10$ solutions
$2P=8,\ \ (Q-P)+R+S=4$ there are $\binom{4-1}{3-1}=3$ solutions
Total $70$ solutions
