Given the set $A=\{1,2,\dotsc,14\}$, find all subsets of $7$ elements that sum to a multiple of $7$.

Consider the collection $F_a$ of all sets $M \subseteq A$ of cardinality $7$ such that $M \cap \{1,\dots 7\}$ has cardinality $a$.

Let $\sigma$ be the permutation of $A$ given by $\sigma(x) = x+1$ for $x = 1, \dots 6$, $\sigma(7) = 1$, $\sigma(x) = x$ for $x = 8, \dots 14$. Then the mapping $M \mapsto \sigma(M)$ is a permutation of $F_a$. Moreover, if the sum of the elements of $M$ is $b$, then the sum of the elements of $\sigma(M)$ is $b + a$ modulo $7$.

Thus the number of elements of $F_a$ with sum $b$ modulo $7$ is the same as the number with sum $b + a$ modulo $7$. When $a \ne 0,7$, repeated application of this argument shows that this number is independent of $b$. Therefore exactly $1/7$ of the elements of $F_a$ have sum $0$ modulo $7$.

$F_0$ and $F_7$ are the only exceptions to this rule, and they each have one element. Therefore the required number is $\frac{1}{7}\left[\binom{14}{7} - 2\right] + 2 = 492$.