Elementary combinatorics problem - why am I wrong?

In summary, the conversation discusses the number of 3-person combinations that can be formed with 6 unique experienced persons and 2 unique inexperienced persons, where at most one person in the group can be inexperienced. The answer is 6*5*4/3! + 6*5*2/2!, which is calculated by considering the order of the experienced persons, but not the inexperienced person. The conversation also explains why the correction factor for the second term should be a 2-permutation instead of a 3-permutation, in order to avoid double counting.
  • #1
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6 unique experienced persons
2 unique inexperienced persons

How many 3-person combinations if at most one in the group can be inexperienced?

6*5*4/3! + 6*5*2/2! is the answer.

My answer was
6*5*4/3! + 6*5*2/3!

Why does the book divide by a 2-permutation and not a 3-permutation for the second term (all combinations consisting of 2 experienced persons and 1 inexperienced person)? My (flawed) rationale is that you have three slots to be filled; we have 6 experienced choices for the first slot, 5 remaining experienced choices for the second slot, and 2 inexperienced choices for the last slot. Since, we don't consider order, we divide by a correction factor of 3 slots permuted.
 
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  • #2
If you have a group consisting of 2 experienced and one inexperienced, there are 6 choose 2 ways of picking the 2 experienced people, and 2 ways to pick the inexperienced one. Try multiplying those out
 
  • #3
Thanks for your answer, but I'm looking for my flaw visually; why am I under-counting in the second term. Why is my last paragraph explanation wrong?
 
  • #4
Your main goal is not to take care of order, you want to take care of double counting. Usually these things are the same, but when dividing out you only want to take care of order within each subset of like objects. So here we have one pair of like objects (the experienced people) whose order is irrelevant, meaning you only divide by 2!

So if the inexperienced people are named Adam and Abe, and some experienced people are Bob, Billy and Bubba:

One possibility is to pick Bob, Bubba and Adam

One possibility is to pick Bubba, Bob and Adam. (so the group of Bob, Bubba and Adam has been counted twice)

Picking Adam, then Bubba then Bob is not a possibility because you said you were picking two experienced people, then one inexperienced person, so you never counted this possibility to begin with. (so the group was only counted twice, not more times)
 
  • #5
That finally cleared it up for me. Thanks a lot.
 

1. Why is it necessary to use a specific formula in combinatorics problems?

Combinatorics deals with counting and arranging objects in a systematic way. Using a formula helps to ensure that all possible combinations are accounted for and avoids errors that may occur with manual counting.

2. What is the difference between permutations and combinations in combinatorics?

Permutations refer to the arrangement of objects in a specific order, while combinations do not consider the order of the objects. In other words, permutations are concerned with the way objects are ordered, while combinations are concerned with the selection of objects regardless of their order.

3. How do I know which formula to use for a specific combinatorics problem?

The formula to use depends on the type of problem. For example, if the problem involves arranging objects in a specific order, then the formula for permutations should be used. If the problem involves selecting objects without considering their order, then the formula for combinations should be used.

4. Can I use combinatorics to solve real-world problems?

Yes, combinatorics can be applied to real-world problems in various fields such as mathematics, computer science, and statistics. It is commonly used in areas such as probability, genetics, and cryptography.

5. What are some common mistakes made when solving combinatorics problems?

Some common mistakes include forgetting to account for all possible combinations, mixing up the formulas for permutations and combinations, and not considering restrictions or conditions given in the problem. It is important to carefully read and understand the problem before attempting to solve it.

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