**From New Scientist #2359, 7th September 2002** [link]

Albion played 18 fixtures in a league in which the result of each fixture was decided by the aggregate scores achieved by the teams over two legs, home and away.
But the rule that “away goals count double” applied in all circumstances;
so a team could win a fixture even though it had scored fewer actual goals than the opposition: a team that drew at home 1-1 and lost away 4-6 would win 9-8 on aggregate after the away goals rule was applied.

In each fixture, Albion scored fewer actual goals over the two legs than the opposition but won on aggregate once the away goals rule was applied.
Even with away goals counting double Albion’s aggregate score in each fixture always remained in single figures. Twelve different scores were involved in Albion’s home legs, six occurring once
and six twice. The scores in the 18 away legs were all different.
(A win by any score is different from a loss by that score).

What were the scores in the away legs of the fixtures in which Albion drew at home 0-0? Give each answer in the form *x*–*y*, with Albion’s score first in each case and without applying the away goals rule.

[enigma1203]

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Solution:The scores in the away legs of the fixtures where Albion drew 0-0 at home were 4-7 and 2-3.The scores in the home and away legs of each fixture as well as the aggregate score for Albion are given in the diagram below:

In fact there are only 18 possible scores for the away legs, so we need to choose a corresponding home leg score for each possible away leg score, and there is only one way to do this such that six of the home leg scores are used once and six of them are used twice. In the diagram the 18 fixtures are grouped by home leg score so it is easy to see how many times each home leg score is used.