(iii) To use 4 different colours, choose 4 colours from the 6 available. This can be done in \(\binom{6}{4}\) ways. For each selection, the 4 pegs can be arranged in \(4!\) ways. Thus, the total number of arrangements is \(4! \times \binom{6}{4} = 360\).

(iv) To use 3 different colours, consider cases where two pegs are of one colour and the other two are of different colours. For example, 2 red, 1 blue, 1 green. The number of arrangements for each case is \(\frac{4!}{2!}\). There are \(\binom{6}{3}\) ways to choose 3 colours from 6. Thus, the total number of arrangements is \(3 \times \frac{4!}{2!} \times \binom{6}{3} = 720\).

(v) Using any of her 12 pegs, consider cases where two pegs are of one colour and two pegs are of another colour. For example, 2 red, 2 blue. The number of arrangements for each case is \(\frac{4!}{2!2!} = 6\). There are \(\binom{6}{2}\) ways to choose 2 colours from 6. Thus, the total number of arrangements is \(6 \times \binom{6}{2} = 90\).

Adding all these possibilities gives the total number of arrangements: \(360 + 720 + 90 = 1170\).