Let A = \ 1, 2, 3, , n\ where n ≥ 5 . Define B = \ T A : either 1 T...

By De Morgan's law, $(B \cup C)' = B' \cap C'$.

The set $B'$ consists of subsets $T$ of $A$ such that $1 \in T$ and $\{2, 3\} \not\subseteq T$. This means $T$ must contain $1$, but cannot contain both $2$ and $3$.
Thus, $T \cap \{1, 2, 3\}$ can be exactly $\{1\}$, $\{1, 2\}$, or $\{1, 3\}$.

The set $C'$ consists of subsets $T$ of $A$ such that the sum of all elements of $T$ is odd.

Let $T = T_1 \cup T_2$, where $T_1 = T \cap \{1, 2, 3\}$ and $T_2 = T \cap \{4, 5, \dots, n\}$.

Case 1: $T_1 = \{1\}$. The sum of elements in $T_1$ is $1$ (odd). For the total sum to be odd, the sum of elements in $T_2$ must be even.
Case 2: $T_1 = \{1, 2\}$. The sum is $3$ (odd). The sum of elements in $T_2$ must be even.
Case 3: $T_1 = \{1, 3\}$. The sum is $4$ (even). The sum of elements in $T_2$ must be odd.

For any non-empty set of consecutive integers starting from $4$ (since $n \ge 5$, the number $5$ is present), exactly half of its subsets have an even sum and half have an odd sum.
The total number of subsets of $\{4, 5, \dots, n\}$ is $2^{n-3}$.
Therefore, the number of subsets $T_2$ with an even sum is $2^{n-4}$, and with an odd sum is $2^{n-4}$.

In each of the $3$ cases for $T_1$, there are $2^{n-4}$ valid choices for $T_2$.
So, $n(B' \cap C') = 3 \times 2^{n-4}$.

We are given $n(B \cup C) = 3328$.
Using $n(B \cup C) = 2^n - n(B' \cap C')$, we get:
$2^n - 3 \times 2^{n-4} = 3328$
$2^{n-4}(2^4 - 3) = 3328$
$13 \times 2^{n-4} = 3328$
$2^{n-4} = 256$
$2^{n-4} = 2^8$
$n - 4 = 8 \implies n = 12$.

Answer: $12$