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Solving problems about selections using combinations

  1. May 20, 2006 #1
    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 dont know how to apply them.:eek:

    help muchos appreciated
     
  2. jcsd
  3. May 20, 2006 #2

    LeonhardEuler

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    Gold Member

    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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