Committee combination math problem

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To form a committee of 3 from 4 lawyers, 1 minister, and 3 retailers, with at least one retailer included, the calculation starts with selecting one retailer, leaving 7 people to choose from. The number of combinations is calculated as 3C(7,2). An alternative approach involves calculating the complement by determining how many committees can be formed without retailers, which is C(5,3), and then subtracting this from the total combinations C(8,3). The final result shows there are 46 valid combinations when accounting for the requirement of including at least one retailer. This problem highlights the importance of understanding combinatorial principles in committee selection scenarios.
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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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