• 69, I think. bbbbgggg bbbgbggg bbbggbgg bbbgggbg bbbggggb bbgbbggg bbgbgbgg bbgbggbg bbgbgggb bbggbbgg bbggbgbg bbggbggb bbgggbgb bbggggbb bgbbbggb bgbbgbgg bgbbggbg bgbbgggb bgbgbbgg bgbgbgbg bgbgbggb bgbggbbg bgbggbgb bgbgggbb bggbbbgg bggbbgbg bggbbggb bggbgbbg bggbgbgb bggbggbb bgggbbbg bgggbbgb bgggbgbb bggggbbb gbbbbggg gbbbgbgg gbbbggbg gbbbgggb gbbgbbgg gbbgbgbg gbbgbggb gbbggbbg gbbggbgb gbbgggbb gbgbbbgg gbgbbgbg gbgbbggb gbgbgbbg gbgbgbgb gbgbggbb gbggbbbg gbggbbgb gbggbgbb gbgggbbb ggbbbbgg ggbbbgbg ggbbbggb ggbbgbbg ggbbgbgb ggbbggbb ggbgbbbg ggbgbbgb ggbgbgbb ggbggbbb gggbbbbg gggbbbgb gggbbgbb gggbgbbb ggggbbbb
  • That depends on what you mean by a "way" of having children. Some of the more obvious answers: One. Four boys and four girls, there's no other way of getting that many boys or that many girls. 40320. Usually each child has its own distinct DNA. The information in DNA can be converted into a number, and there are 8-factorial = 8! = 40320 orderings for eight numbers. 70. If each boy is considered to be the same until it is born and then named, so that the first four boys will always be named the same way and the same for the girls, then you can find the number of ways using binomial coefficients. The number of ways of arranging four equal boys and four equal girls into an order, is the same as the number of ways of choosing four numbers from 1 to 8 representing the position of the boys in that order. The number of ways of choosing a set of 4 numbers from 8 is called "8 choose 4" sometimes written 8C4 and is calculated as (8*7*6*5)/(4*3*2*1) (The numererator product in nCr stops when you have r numbers) (8*7*6*5)/(4*3*2*1) = 70
  • Only one way
  • Uh, one?
  • There is only one way to have boys and girls. If you meant something else, you need to clarify the question to suggest a particular answer.

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