Exercise 1.3.7 in Grimmett & Stirzaker's 'Probability and Random Processes' I'm having trouble solving exercise 1.3.7 (and even understanding the solution) from Grimmett & Stirzaker's Probability and Random Processes, which reads:

You are given that at least one, but no more than three, of the events $A_r$, $1 \le r \le n$, occur, where $n \ge 3$. The probability of atleast two occurring is $1/2$. If P$(A_r)$ = $p$, P$(A_r \cap A_s)$ = $q$, $r \ne s$ and P$(A_r \cap A_s \cap A_t) = x$, $r \lt s \lt t$ show that $p \ge 3/(2n)$, and $q \le 4/n$.

I'm linking the solution to the problem here. I am not able to understand what the ﬁrst three terms in the latter series (in the solution) are and how they lead to the final equation.
The second term in the series seems to be the intersection of all the 2-pairs according to me.
 A: The feedback from Lord Shark the Unknown helped but the following way of counting drove the point across:
For the second term in the sequence, we are essentially choosing 3 numbers (a,b,c) in $n\choose{3}$ for $$A_{r,s} \cap A_{t,u} = A_a \cap A_b \cap A_c$$ such that $$a \lt b \lt c \\ r \lt s, t \lt u \\ (r,s) \ne (t,u)$$.
Now, given the constraints, the following 3 can be the possible orderings of a,b,c:
$$\{\{(a,b),(b,c)\},\{(a,b),(a,c)\},\{(a,c),(b,c)\}\}$$
For the third term, we still have to choose 3 distinct numbers (a,b,c) in $n\choose{3}$ ways, but now $$A_{r,s} \cap A_{t,u} \cap A_{v,w} = A_a \cap A_b \cap A_c$$ such that $$a \lt b \lt c \\ r \lt s, t \lt u, v \lt w \\ (r,s) \ne (t,u) \ne (v,w)$$
Now, given the constraints, only the following can be the possible ordering of a,b,c:
$$\{(a,b),(b,c),(a,c)\}$$
Therefore, the second term contributes $3 {n\choose{3}}$ and the third term contributes ${n\choose{3}}$.
Thanks again for the feedback.
A: This is about the inclusion-exclusion principle. The linked solution
applies it to the events $A_{r,s}=A_r\cap A_s$. The solution displays
these events and their double intersections $A_{r,s}\cap A_{t,u}$.
The third term is the sum of their triple intersections. One gets
nonzero contributions from triple intersections like
$$A_{r,s}\cap A_{r,t}\cap A_{s,t}=A_r\cap A_s\cap A_t.$$
There are $\binom{n}3$ of these. But the other terms  in this
 sum involve triple intersections which equal the intersection of
four or more of the $A_i$, so their probability is zero.
Likewise in the second sum, only intersections like $A_{r,s}\cap A_{r,t}$
count.
