Committee combination math problem

In summary, there are 46 possible ways to choose a committee with 3 out of 4 lawyers, 1 minister, and 3 retailers, where one person must be a retailer. This is calculated by subtracting the number of committees without any retailers (C(5,3)) from the total number of committees (C(8,3)).
  • #1
PiRsq
112
0
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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  • #2
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?
 
  • #3
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:
 

FAQ: Committee combination math problem

What is a committee combination math problem?

A committee combination math problem is a type of problem where you are given a set of objects or individuals and are asked to find all possible combinations or arrangements of those objects or individuals.

How do you solve a committee combination math problem?

To solve a committee combination math problem, you can use a mathematical formula or approach called combination. The formula for combination is nCr = n! / r!(n-r)!, where n is the total number of objects or individuals and r is the number of objects or individuals to be chosen for the committee.

What are the steps to solve a committee combination math problem?

The steps to solve a committee combination math problem are:

  • Determine the total number of objects or individuals (n).
  • Determine the number of objects or individuals to be chosen for the committee (r).
  • Apply the combination formula, nCr = n! / r!(n-r)!, to find the total number of combinations.
  • Use the result to list out all possible combinations.

Can a committee combination math problem have repeated objects or individuals?

No, a committee combination math problem does not allow for repeated objects or individuals. Each combination must have unique objects or individuals.

How is a committee combination math problem different from a permutation problem?

A committee combination math problem deals with selecting objects or individuals without considering the order, while a permutation problem involves arranging objects or individuals in a specific order. In a committee combination math problem, the order of the chosen objects or individuals does not matter, but in a permutation problem, the order matters.

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