Problem (Combinatorics):
Let $T=T N Y W R$. A group of $n$ friends goes camping; two of them are selected to set up the campsite when they arrive and two others are selected to take down the campsite the next day. Compute the smallest possible value of $n$ such that there are at least $T$ ways of selecting the four helpers.

Solution:
There are $\left(\begin{array}{l}n \\ 2\end{array}\right)$ ways of choosing the two people to set up and $\left(\begin{array}{c}n-2 \\ 2\end{array}\right)$ ways of choosing the two people to take down the campsite, so there are $\frac{n(n-1)}{2} \cdot \frac{(n-2)(n-3)}{2}$ ways of choosing the four people, or $\frac{n(n-1)(n-2)(n-3)}{4}$ ways total; call this function $C(n)$. For the least $n$ such that $\frac{n(n-1)(n-2)(n-3)}{4} \geq T$, as a rough approximation, note that $n-3<\sqrt[4]{4 T}<n$. With $T=184$, the approximation becomes $n-3<\sqrt[4]{736}<n$. Now $5^{4}=625$ while $6^{4}=1296$, so $5<n<9$. Try values starting from $n=6$ :

$$
\begin{aligned}
& C(6)=\frac{6 \cdot 5 \cdot 4 \cdot 3}{4}=90 \\
& C(7)=\frac{7 \cdot 6 \cdot 5 \cdot 4}{4}=210 .
\end{aligned}
$$

Thus $n=7$.

Final answer: 7