Solving problems about selections using combinations

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SUMMARY

This discussion focuses on solving selection problems using combinations and permutations, specifically in the context of forming a team of 3 men and 2 women from a group of 6 men and 5 women. The key takeaway is the distinction between permutations, where order matters, and combinations, where it does not. The formula for combinations is derived as 12C3 = 12! / (3! * (12-3)!), which accounts for the selection of team members without regard to order. The discussion emphasizes the importance of understanding these concepts to effectively tackle similar problems.

PREREQUISITES
  • Understanding of permutations and combinations
  • Familiarity with factorial notation
  • Basic knowledge of combinatorial mathematics
  • Ability to apply mathematical reasoning to real-world scenarios
NEXT STEPS
  • Study the concept of factorials and their applications in combinatorial problems
  • Learn how to derive and use the combinations formula in various contexts
  • Explore advanced combinatorial techniques such as the binomial theorem
  • Practice solving real-world problems involving selections and combinations
USEFUL FOR

Students, educators, and professionals in mathematics, statistics, or any field requiring combinatorial analysis will benefit from this discussion.

RigidBody
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hello

i need help solving problems involving selecting things. like for example find the number of ways in which a team of 3 men and 2 women can be selected from a group of 6 men and 5 women.

:bugeye: i know how to do perms and combs but just don't know how to apply them.:eek:

help muchos appreciated
 
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Alright, first I'll explain when to use permutations so that combinations make more sense. Suppose you want to know how many 5 letter words there are with no letter repeated. You can think of choosing the letters in order. For the first letter there would be 26 choices. For the second there would only be 25 because you can't use the first letter again. And so on. The number of possibilities is 26*25*24*23*22=[itex]\frac{26!}{21!}[/itex]= 26P5. Use permutations when you are counting the number of ways to choose objects without replacement where the order is important.

Now suppose the order is not important, as when choosing the members of a team. Suppose you are choosing a team of 3 people from a group of twelve. We already figured out that if the order did matter then the number of possibilities would be 12P3. But the order doesn't matter. How do we account for that? Suppose we have a team A,B,C. This team is the same as B,C,A or A,C,B and so on. From what I said in the first paragraph, its clear that the number of teams consisting of A,B and C is 3P3=[itex]\frac{3!}{(3-3)!}=\frac{3!}{0!}=3![/itex]. So each team is counted 3!=6 times. This means that the real number of teams is only one sixth of what we counted. So the real number of teams where the order does not matter is 12P3/3!=[tex]\frac{\frac{12!}{(12-3)!}}{3!}=\frac{12!}{(12-3)!3!}[/tex]=12C3. Use combinations when choosing things without replacement when order does not matter.

Finally, coming to your example. Basically you can think of what you are doing as choosing two teams: one of women and one of men. You can use the method I just decribed to find the number of ways each of those teams could be chosen. Then how do you find the total number of ways the combined team could be chosen? See if you can reason it out.
 

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