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    Can anybody help me do this probability question on permutations? :confused: It's been driving me nuts...

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    (Original post by doctor_b)
    Can anybody help me do this probability question on permutations? :confused: It's been driving me nuts...
    I remember doing this question and getting a neatish solutioin, but can't remember how. However, I've worked out a different method which gets the answer.

    Split it into three different scenarios:
    1) First three are all different colours
    No. of ways = 6*5*4*6 (because the first hole can be any of one of 6 colours, the second must be different so it can be any one of 5 colours, etc., but the last hole does not matter)

    2) First two are same colour
    No. of ways = 6*1*5*5 (second one must match first, leaving only 5 different colours for the last two pegs)

    3) First two are different, but third is the same as one of previous two
    No. ways = 6*5*2*5

    Total = 6*5*4*6 + 6*1*5*5 + 6*5*2*5 = 1170

    Another method is to split it up according to the number of colours used.

    4 colours
    6*5*4*3 [choose the colour of the first hole, choose the colour of second hole, choose the colour of the third hole, choose the colour of the fourth hole]

    3 colours (one pair of similar pegs plus two pegs of other colours)
    6*4C2*5*4 [choose the colour of the similar pair, choose where the pair goes, choose the colours of the other two holes]

    2 colours (two pairs of similar pegs)
    6C2*4C2 [choose the colours of the pairs, choose where they go]

    = 6*5*4*3 + 6*4C2*5*4 + 6C2*4C2
    = 1170
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Updated: December 11, 2005

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