Committee combination math problem

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SUMMARY

The committee combination problem involves selecting 3 members from a group of 4 lawyers, 1 minister, and 3 retailers, with the stipulation that at least one member must be a retailer. The correct approach is to first select 1 retailer (3 choices) and then choose 2 additional members from the remaining 7 individuals, calculated as C(7,2). Alternatively, the problem can be solved using complementary counting, determining the total combinations without retailers (C(5,3)) and subtracting from the total combinations (C(8,3)), resulting in 46 valid combinations.

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For a commitee, 3 people out of 4 lawyers, 1 minister and 3 retailers are to be chosen. 1 person in the commitee must be a retailer, how many ways are there to choose the commitee?


1st: There must be a retailer thus we have 3 choices
2nd: Out of the remaining 7 people, the possible combinations are C(7,2)

So there must be 3C(7,2) possibilities. Is it right?
 
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How many times did you count the cases with two retailers?

This problem is probably easier to solve by first solving its complement: how many committees don't have any retailers?
 
Then you would have 5 people remaining. I would have C(5,3). That is if there were no retailers. With retailers, I would have, C(8,3). Thus C(8,3)-C(5,3) equals 46...OMG I WISH I ONLY KNEW THAT WHEN I WAS TAKING THE TEST! I NEED MORE PRACTICE, thanks Hurkyl!:smile:
 

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