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Nov 2010 p61 q6
2869
Pegs are to be placed in the four holes shown, one in each hole. The pegs come in different colours and pegs of the same colour are identical.
Beryl has 12 pegs consisting of 2 red, 2 blue, 2 green, 2 orange, 2 yellow and 2 black pegs. Calculate how many different arrangements of coloured pegs in the 4 holes Beryl can make using
4 different colours,
3 different colours,
any of her 12 pegs.
Solution
(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: \(360 + 720 + 90 = 1170\).
