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Let's say three couples are selected to go to a spacebase in Mars, and there is enough food and air in the base to survive for a couple of hundred years.Each couple has one son and one daughter.Find the number of ways each of the sons and daughters of the couples can marry when they grow up.

There are a lot of potential tricks here, especially since you don't supply the ages. We could have homosexual, bigamous, serial and intergenerational marriages. But if we play it straight and assume that each son has to marry a daughter who is not his sister, there are only two possibilities. Son A can marry Daughter B or Daughter C. The brother of the bride must marry the other Daughter B or C. Daughter A must marry the remaining son.

you're right Aaron, i'm sorry but let's suppose this is a fairly simple world where marriage between same gender does not exist, either between parents and sons or intergenerational in general, and the couples remain married.

anybody can marry anyone, only 6 types of childeren anyway. You assume "spacebase in Mars" so then the natural answear to multiplication is CLONES!!!

Is there a trick in this question?denote A, B and C the children from each of the 3 couples, and let 1 and 0 parameetrize their gender.then you have the following five possibilities,A0 + B1, A1 + B0, C1 and C0 do not get married,C0 + B1, B0 + C0, A1 and A0 do not get married,C0 + A1, A1 + C0, B1 and B0 do not get married,plus the two possibilities mentioned by Aaron.

I interpreted the question as "each son has to marry a daughter." If you relax that and allow people not to marry then both of my solutions exist in 8 variants: all three couples marry (1), two couples marry (3), one couple marries (3) and no couples marry (1). Only the last one is identical for the two solutions, so you have a total of 15 distinct solutions.