If $n(A) = 3$,$n(B) = 6$ and $A \subseteq B$,then the number of elements in $A \cup B$ is equal to

If the number of elements in a set $A$ is $n(A) = 40$,$n(B) = 26$,and $n(A \cap B) = 16$,then what is the value of $n(A \cup B)$?

$A$ set contains $2n + 1$ elements. The number of subsets of this set containing more than $n$ elements is equal to

$2n (A \setminus B) = n (B \setminus A)$ and $5n (A \cap B) = n (A) + 3n (B)$,where $P \setminus Q = P \cap Q^C$. If $n (A \cup B) \leq 10$,then the value of $\frac{n(A) \cdot n(B) \cdot n(A \cap B)}{8}$ is:

Find the cardinal number of the following set ${x: x = 2n, n \in N, 4 \leq x \leq 11}$.

If $A = \{x : x = \frac{n-1}{n+1}, n \in W\}$ and $\{n \leq 10\}$,point out the correct statement from the following: