a) Freddie has 6 toys cars and 3 toy buses, all different. i) Freddie arranges these 9 toys in a line. Find the number of possible arrangements · if there is a car at each end of the line and no buses are next to each other.

$Car...Car...Car...Car...Car...Car$

Six cars we can arrange in $6!=720$ ways.

Buses can be filled in the gap between two toy cars.

There are five gaps between the cars.

Three buses can be filled in five gaps in $P^5_3=\dfrac{5!}{(5-3)!}=60$ ways.

So, the total number of ways to arrange is

$720(60)=43200$