What is the flaw in this approach? $12$ 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.
My method:
I first selected one from each city which can be done in $4 \times 4 \times 4=64$ ways and then remaining $3$ can be chosen from the left over $9$ in $\binom{9}{3}=84$ ways. So the total ways is $64 \times 84=5376$. I know its wrong but where did I go wrong?
 A: You are overcounting, as the order in which delegates are chosen from a single city does not matter.

To obtain the correct answer, notice that you are counting outcomes where the six delegates do not all come from exactly two cities.
A: As indicated in the comments, the issue is that you're counting each unordered group of delegates multiple times based on whether they're selected in the first part of your process (the guaranteed representative for each city) or the second part. (filling the rest of the roster)
I think in this case it's easier to count the number of ways that we can select groups which do not fit our criteria - how many committees of $6$ members can we make which exclude one of the cities entirely? If you'd like to try I think this hint would likely lead you in the correct direction, but I've included my work below:

We can consider this as the union of three sets: let's say $A$ is the set of committees which exclude $C_1$, $B$ is the set of those which exclude $C_2$, and $C$ is the set of those which exclude $C_3$. A key insight here is that these sets are disjoint: for a committee to exclude two cities simultaneously, it would have to draw $6$ members from the remaining city, which only has $4$ delegates. So, we simply have that
$$|A \cup B \cup C| = |A| + |B| + |C|$$
and because all three sets are the set of all possible combinations of $6$ delegates being taken out of a possible $8,$ we have that
$$|A| = |B| = |C| = \binom{8}{6}$$
Now, our original set is the set of committees which are not in $A, B,$ or $C$: the set $\overline{A} \cap \overline{B} \cap \overline{C} = \overline{(A \cup B \cup C)}.$ So, in order to count this set, we can just subtract $|A \cup B \cup C|$ from the total number of committees of $6$ members taken from the $12$ delegates, which is to say $\binom{12}{6}.$ So, our final answer is
$$\bigl|\overline{A} \cap \overline{B} \cap \overline{C}\bigr| = \binom{12}{6} - 3\binom{8}{6} = 840$$
A: You can only exclude any $1$  city while choosing $6$ members, thus
[unrestricted selections] - [selections excluding any one city]
$= \binom{12}6 - \binom31\binom86 = 840$
The flaw in your approach is that you are overcounting by fractionating the choosing process
A: 
$12$ 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:
$$\binom{4}{2} \times \binom{4}{2} \times \binom{4}{2} = 216.$$

$\underline{\text{Final Compuation:}}$
$$48 + 576 + 216 = 840.$$
