your answers.
(i) 6 pegs, all of different colours: 6! (6 factorial) = 720 different arrangements. This is because there are 6 pegs and each peg can be placed in any of the 4 holes, giving 6x6x6x6 = 6! possible arrangements.
(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg: 4! (4 factorial) = 24 different arrangements. This is because there are 4 pegs and each peg can be placed in any of the 4 holes, giving 4x4x4x4 = 4! possible arrangements.
(iii) 4 different colours: 4! (4 factorial) = 24 different arrangements. This is because there are 4 different colours and each colour can be placed in any of the 4 holes, giving 4x4x4x4 = 4! possible arrangements.
(iv) 3 different colours: 3! (3 factorial) = 6 different arrangements. This is because there are 3 different colours and each colour can be placed in any of the 4 holes, giving 3x3x3x3 = 3! possible arrangements.
(v) any of her 12 pegs: 12! (12 factorial) = 479,001,600 different arrangements. This is because there are 12 pegs and each peg can be placed in any of the 4 holes, giving 12x12x12x12 = 12! possible arrangements.
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. Calculate how many different arrangements of coloured pegs in the four holes can be made using
(i) 6 pegs, all of different colours, [1]
(ii) 4 pegs consisting of 2 blue pegs, 1 orange peg and 1 yellow peg. [1]
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
(iii) 4 different colours, [1]
(iv) 3 different colours, [3]
(v) any of her 12 pegs.
Please explain
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1 answer