A small town has a rectangular grid of streets, shown in the figure…

(a) A route is fixed by the places of its $4$ letters N among its $11$ letters: $\binom{11}{4} = 330$ routes.

(b) A route through $P$ goes from $A$ to $P$ ($3$ letters E and $2$ letters N, $\binom{5}{2} = 10$ ways) and then from $P$ to $B$ ($4$ E and $2$ N, $\binom{6}{2} = 15$ ways). So $10 \times 15 = 150$ routes pass through $P$, and $330 - 150 = 180$ avoid it.

(c) A route with $t$ turns has $t + 1$ runs, and the runs alternate between E and N. With $4$ turns there are $5$ runs. If the first run is E, the runs are E, N, E, N, E: three runs of E and two runs of N. Cutting the $7$ letters E into $3$ non-empty runs means choosing $2$ of the $6$ gaps between them, $\binom{6}{2} = 15$ ways. Cutting the $4$ letters N into $2$ runs means choosing $1$ of $3$ gaps, $3$ ways. That gives $15 \times 3 = 45$ routes. If the first run is N, there are three runs of N and two runs of E: $\binom{3}{2}\binom{6}{1} = 18$ routes. In all, $45 + 18 = 63$.

(d) The runs alternate, so the numbers of E runs and N runs differ by at most $1$. There are only $4$ letters N, so at most $4$ runs of N, and then at most $5$ runs of E. That is at most $9$ runs, or $8$ turns. Exactly $8$ turns needs $5$ runs of E and $4$ runs of N, so the route starts and ends with E and every N stands alone. The $7$ letters E are cut into $5$ runs in $\binom{6}{4} = 15$ ways, and the N runs are forced. So the greatest number of turns is $8$, made by $15$ routes.

(e) Proof. A route with exactly $2k$ turns has $2k + 1$ runs. This number is odd, so the first run and the last run have the same letter.

Case 1: the route starts with E. Then it has $k + 1$ runs of E and $k$ runs of N. The lengths of the E runs are $k + 1$ positive whole numbers that add up to $7$, in order. Such a list is fixed by choosing which $k$ of the $6$ gaps between the $7$ letters E are cuts, so there are $\binom{6}{k}$ lists. In the same way the lengths of the N runs can be chosen in $\binom{3}{k-1}$ ways. Once both lists are chosen the word is fixed, because the runs alternate and start with E. So Case 1 gives $\binom{6}{k}\binom{3}{k-1}$ routes.

Case 2: the route starts with N. Now there are $k + 1$ runs of N and $k$ runs of E, which gives $\binom{3}{k}\binom{6}{k-1}$ routes.

Every route with $2k$ turns is in exactly one case, so the total is the sum of the two terms. Both terms are needed because a route may start in either direction, and the two directions have different numbers of letters ($7$ and $4$), so the two cases do not give the same count. For $k = 2$ the formula gives $45 + 18 = 63$, as in (c).