$12$ delegates exists in three cities $C_1,C_2,C_3$ each city having $4$ delegates. A committee of six members is to be formed from these $12$ such that at least one member should be there from each city.
First, see my comment, immediately following the posted question. There, I recommend Inclusion-Exclusion, which is the method used by all of the other answers. Although the direct approach is definitely the inferior approach to this problem, it is workable. This answer used the direct approach.
Let $x_1, x_2, x_3,$ denote the number of members contributed by each of $C_1, C_2, C_3,$ respectively. There are only three distinct ways that positive integer values $~~\leq 4~~$ can be assigned to the variables $x_1, x_2, x_3$, such that $x_1 + x_2 + x_3 = 6.$ The distinct ways are:
- Case 1:
One of $x_1, x_2, x_3$ equals $(4)$, and the other two variables equal $(1)$.
- Case 2:
One of $x_1, x_2, x_3$ equals $(3)$, one of the remaining two variables equals $(2)$, and the remaining variable equals $(1)$.
- Case 3:
Each of $x_1, x_2, x_3$ equals $(2)$.
In effect, you are partitioning the number $(6)$ as the sum of three positive integers, with each integer $~~\leq 4$.
The remainder of this answer provides the corresponding explicit calculations for this inferior direct approach.
Note
In general, for $~k \in \{1,2,3,4\},~$ the number of ways of selecting $k$ delegates from a city that has $4$ delegates is $~\displaystyle \binom{4}{k}.$
$\underline{\text{Case 1:}}$
One of $x_1, x_2, x_3$ equals $(4)$, and the other two variables equal $(1)$.
There are three choices for which variable will be assigned the value $(4)$.
Therefore, the enumeration is:
$$3 \times \binom{4}{4} \times \binom{4}{1} \times \binom{4}{1} = 48.$$
$\underline{\text{Case 2:}}$
One of $x_1, x_2, x_3$ equals $(3)$, one of the remaining two variables equals $(2)$, and the remaining variable equals $(1)$.
There are three choices for which variable will be assigned the value $(3)$. Then, there are two choices for which of the remaining variables will be assigned the number $(2)$.
Therefore, the enumeration is:
$$(3 \times 2) \times \binom{4}{3} \times \binom{4}{2} \times \binom{4}{1} = 576.$$
$\underline{\text{Case 3:}}$
Each of $x_1, x_2, x_3$ equals $(2)$.
The enumeration is:
$$\binom{4}{2} \times \binom{4}{2} \times \binom{4}{2} = 216.$$
$\underline{\text{Final Compuation:}}$
$$48 + 576 + 216 = 840.$$