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# Tough probability watch

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

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2. (Original post by doctor_b)
Can anybody help me do this probability question on permutations? 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
3. 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]

Total
= 6*5*4*3 + 6*4C2*5*4 + 6C2*4C2
= 1170

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Updated: December 11, 2005
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