Why is the number of bee routes between $5$ flowers not equal to $5!$ On a BBC program about algorithms, they talked about solving the travelling salesman problem. They used the analogy of a bee hive with five flowers to visit before returning to the hive? They said that there were 60 possible routes. Am I missing something or should it not be 5! = 120?
You have not missed anything.

Including the beehive, there are $6$ nodes, and the total number of paths for $n$ nodes is given by $(n-1)!$, so it has to be $120$ for this case.

See under "Difficulty" here

Actually, the real difficulty of this type of problem is finding the shortest route, and trying out all possible routes becomes prohibitive as $n$ increases.

PS:

We have to count number of paths from hive (H) to flowers and back
For 3 flowers, say, [ mind it, $n = 4,$ possible routes are

$H-1-2-3-H,\; H-1-3-2-H,\; H-2-1-3-H,$

$H-2-3-1-H, \;H-3-1-2-H,\; H-3-2-1-H,$

$6$ or $3! = (n-1)!$ 

PPS:

So the number of routes with $5$ flowers = $120$, 

but the number of routes of the same length (assuming reverse route doesn't change distance ) = $60$

but the question as given here didn't ask that.
You have not missed anything.

Including the beehive, there are $6$ nodes, and the total number of paths for $n$ nodes is given by $(n-1)!$, so it has to be $120$ for this case.

See under "Difficulty" here

Actually, the real difficulty of this type of problem is finding the shortest route, and trying out all possible routes becomes prohibitive as $n$ increases.

PS:

We have to count number of paths from hive (H) to flowers and back
For 3 flowers, say, [ mind it, $n = 4,$ possible routes are

$H-1-2-3-H,\; H-1-3-2-H,\; H-2-1-3-H,$

$H-2-3-1-H, \;H-3-1-2-H,\; H-3-2-1-H,$

$6$ or $3! = (n-1)!$ 

PPS:

So the number of routes with $5$ flowers = $120$, 

but the number of routes of the same length (assuming reverse route doesn't change distance ) = $60$

but the question as given here didn't ask that.