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Originally posted by 1234567
question:
a committe of 3 is choosen from a group of four males and five females. the committe must include at least 1 male and at least 1 female. how many combinations are possible/
in this question, the order doesnt' matter.
from males
4c1, females 5c1, and 7c1 from the rest of the them
then the answer is 4c1 x 5c1 x 7c1 = 140 in the same manner as the question above
but the model answer the paper gave is
4c1 x 5c2 + 4c2x 5c1.
the concept is sort for the same, but the answer is different, they get 70........
that's why it got me confused...
It is the same, that is why the 140 is double the answer.Originally posted by Dumbarse
"if you take 4c1*5c1*7c1, you'll make choosing woman A (in the 5) then woman B (in the rest 7 ) AND woman B then woman A two different ways. Which is not good. "
isnt it the same for men but??
why is the answer
ways (2 women) + ways (1 woman)
I got 270 as the answer too. Some sydney boys teacher has some real competence problems.Originally posted by wogboy
I'm quite sure that 270 should be the answer. Gee I hate 3U probability, perms, and combos!
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