We were propping up the bar in the Jolly Quizmaster Arms, and chatting to the landlord. He started to tell us about the 100 Club Christmas Lunch that he was organising.
He told us that of the 100 members, 90 liked red cabbage, 80 liked parsnips and 70 liked sprouts, and that he had to plate up the correct vegetables for each member.
We sympathised with his difficulty, and out of idle curiosity asked if any members didn't eat any vegetables. He replied that, commendably, very few did. And then knowing we like a puzzle he added cryptically:
19 times as many members eat all three vegetables as those that eat none, and the only people who eat only one vegetable all eat red cabbage.
Well that spoiled the evening, as we could not leave without working out how many people had three vegetables, two vegetables and one vegetable, and which do they eat.
If the ten members who don't eat red cabbage all eat parsnips there must be 70 people who eat both. If the 30 who don't eat red cabbage and parsnips all eat sprouts there must be at least 40 members who eat all three vegetables. The number who eat all three can't be more than 70 - the number who eat sprouts.
So we need to find a number divisible by 19 between 40 and 70, and 57 is the only one. There are 57 members who eat all three vegetables and only three who eat none.
Of the remaining 40 some eat one vegetable (red cabbage) and some eat two.
Let a = the number who eat red cabbage and parsnips, b = the number who eat red cabbage and sprouts, c = the number that eat parsnips and sprouts, and d = the number who eat red cabbage only.
$$ a + b + c + d = 40$$
Red cabbage eaters
$$ 57 + a + b + d = 90 $$
$$ a + b + d = 33 $$
$$ 57 + a + c = 80 $$
$$ a + c = 23 $$
$$ 57 + b + c = 70 $$
$$ b + c = 13 $$
And from these simultaneous equations
a = 16, b = 6, c = 7 and d = 11.
So 57 members eat all three vegetables, 16 eat red cabbage and parsnips, six eat red cabbage and sprouts, 7 eat parsnips and sprouts, eleven eat only red cabbage, and three eat no vegetables at all.