Let $S = \{1, 2, 3, \ldots, 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by $5$. What is the maximum number of elements possible in $A$?

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $B = \{3, 6, 7, 9\}$. Then the number of elements in the set $\{ C \subseteq A : C \cap B \neq \phi \}$ is

Let $A_1, A_2, \ldots, A_m$ be non-empty subsets of $\{1, 2, 3, \ldots, 100\}$ satisfying the following conditions:
$1.$ The numbers $|A_1|, |A_2|, \ldots, |A_m|$ are distinct.
$2.$ $A_1, A_2, \ldots, A_m$ are pairwise disjoint.
(Here $|A|$ denotes the number of elements in the set $A$).
Then,the maximum possible value of $m$ is:

$A$ and $B$ are non-singleton sets and $n(A \times B) = 35$. If $B \subset A$,then ${}^{n(A)}C_{n(B)}$ is equal to

In a class of $35$ students,$24$ like to play cricket and $16$ like to play football. Also,each student likes to play at least one of the two games. How many students like to play both cricket and football?

$A$ survey shows that $73 \%$ of the persons working in an office like coffee,whereas $65 \%$ like tea. If $x$ denotes the percentage of them who like both coffee and tea,then $x$ cannot be