Choose the child to get on first any of the 4 children besides Jenny.
For each of those 4 choices for the child to get on first, there are 3
ways to choose the child to get on second as any of 3 the children
other than Jenny and the child chosen to get on first.
That's 4*3 or 12 ways to choose the first two children to get on the bus.

For each of those 12 ways to choose the two children to get on first and 
second, there is only 1 way to choose the child that gets on third, for 
that can only be Jenny.

That's 4*3*1 = 12 ways to choose the first 3 children to get on the bus
first second and third.

For each of those 12 ways to choose the first 3 children to get on the
bus there are 2 ways to choose the child that gets on fourth, the two
remaining children.

That's 4*3*1*2 = 24 ways to choose the first 4 children to get on the bus.

That leave only 1 child to pick to get on the bus last or fifth. 

For each of the 24 ways to pick the first 4 children to get on the bus,
there is only 1 way to pick the last one.

That's 4*3*1*2*1 = 24 ways to choose 5 children to get on the bus in order,
with Jenny always getting on third.

Answer: 24 = 4!

Edwin