<s>[INST] Finding the minimum number of students 
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?

It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
ADDED I am adding the given answer options:
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
 [/INST] For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.
Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly above the main ‘diagonal’. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \le k \le p$ let $$\begin{align*}C_k &= \{\langle m,n \rangle \in S:m=k \lor n=k\}\\ &= \{\langle k,n \rangle:k+1 \le n \le q+1\} \cup \{\langle m,k \rangle:1 \le m \le k-1\};\end{align*}$$ clearly $\vert C_k \vert = q+1-k+k-1=q$, and if $1 \le i < k \le p$, $C_i \cap C_k = \{\langle i,k \rangle\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\vert S \vert$.
Columns $2$ through $p$ of the grid contain $\sum_{k=1}^{p-1} k = \binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so $$\begin{align*}\vert S \vert &= \binom{p}{2} + p(q+1-p)\\
&= \frac{p^2 - p}{2} + pq + p - p^2\\
&= pq - \frac{p^2-p}{2}\\&= pq - \binom{p}{2}.
\end{align*}$$
When $p > q+1$ the same approach works, but it’s no longer possible to get every pair of committees to overlap. First form $\left\lfloor \frac{p}{q+1}\right\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\binom{q+1}{2}=\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2}$ students. The remaining $r = p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor$ committees will then require another $rq - \binom{r}{2}$ students, for a grand total of $$\begin{align*}
\vert S \vert &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + rq - \binom{r}{2}\\ &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + \left(p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor\right)q - \binom{r}{2}\\
&= pq - \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} - \binom{r}{2}
\end{align*},$$ which does not appear to simplify greatly.
I see that this is essentially Alex’s solution, but expressed a little more concretely.
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