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Also 1 win, 6 drawn and 1 lost
Mike, I don't think you can have this for C because you would then have 3 more wins to distribute between D and E and, because you can't have 2 and 1 for D and E (because that would be repeating the number of wins that C has had), it would have to be 3 for D and 0 for E. This would leave D drawing once and E drawing twice, making altogether 4 draws among AD and E so with whom was C drawing 6 times?
Your alternative seems fine which makes me wonder whether mine is wrong unless there are 2 answers.
Nigs, it took me about 10 minutes to solve. Once I had worked out what A and B had to be, I just played around with the numbers until I found something that worked for all the conditions. That said, now Mike has another solution I am wondering if mine is right.