Freddie has 6 toy cars and 3toy buses, all different. He chooses 4 toys to take in holiday with him?

The statement that "He chooses 4 toys to take in holiday with him" is redundant.

C: stands for a car
B: stands for a bus

Firstly, arrange the 6 cars as the C_C_C_C_C_C (₆P₆).  "_" stands for the space between two cars.

Then, place the 3 buses in 3 spaces out of the 5 spaces (₅P₃).  This ensures that there is a car at each end of the line and no buses are next to each other.

No. of arrangements
= ₆P₆ × ₅P₃
= 6! × [5! / (5 - 3)!]
= 6! × (5! / 2!)
= 6! × (5× 4 × 3! / 2!)
= 6! × (10 × 3!)
= 6! 3! 10
= 43200

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There are many steps because I want to show you that it is 6! × 3! 10. The steps could be simplified as:

No. of arrangements
= ₆P₆ × ₅P₃
= 6! × [5!/2!]
= 43200

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