Problem (Combinatorics):
Let $T=210$. At the Westward House of Supper ("WHS"), a dinner special consists of an appetizer, an entrée, and dessert. There are 7 different appetizers and $K$ different entrées that a guest could order. There are 2 dessert choices, but ordering dessert is optional. Given that there are $T$ possible different orders that could be placed at the WHS, compute $K$.

Solution:
Because dessert is optional, there are effectively $2+1=3$ dessert choices. Hence, by the Multiplication Principle, it follows that $T=7 \cdot K \cdot 3$, thus $K=\frac{T}{21}$. With $T=210$, the answer is 10 .

Final answer: 10