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Combination is not my strong point

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data

    A small company gives bonuses to their employees at the end of the year. 15 employees are entitled to receive these bonuses of whom 7 employees will receive 100$ bonus, 3 will receive 1000$ bonus and the rest will receive 3000$ bonus. In how many possible ways these bonuses can be distributed?

    2. Relevant equations

    nCr = n! / (r!(n-r)!)

    3. The attempt at a solution

    But I am unsure if this correct.
    Would it simply be:
    15C7 x 8C3 x 5C5
     
  2. jcsd
  3. Aug 27, 2014 #2

    Ray Vickson

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    Instead of "guessing", try to think it out logically. In how many ways can you distribute 7 bonuses of $100 and 8 "others" among the fifteen? After that....what?
     
  4. Aug 27, 2014 #3
    My problem when it comes to Combination and Permutation is that I have issue seeing/creating the logic.

    Here was my logic when creating that formula.
    1. Out of 15 people, 7 will receive $100.
    2. Since 7 out of 15 people will get $100, 8 candidates are left and 3 will receive $1000.
    3. lastly, there are 5 left over and since the order does not matter it will just equate to 1 (5C5)

    Am I correct?
     
  5. Aug 27, 2014 #4
    I was going to reply to your comments but decided to have a more in depth think about it as I am having a massive headache thinking about it.

    I just want to say thanks and hopefully when I get my thought together, I would have solved it.
     
  6. Aug 28, 2014 #5

    Orodruin

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    Just to try to offer a different approach that will end with the same result:

    Let us say that we give out the bonuses one by one. How many orders of giving out bonuses exist?
    For each given order, how many orders are equivalent in terms of how much bonus everyone gets?

    How do the two numbers above relate to the number of different ways of distributing the bonuses?
     
  7. Aug 28, 2014 #6
    The total of number of ways the bonuses can be distributed = 1515
    So does this mean that "How many orders of giving out bonuses exist?" = 15?
    Or does 'How many orders of giving out bonuses exist' = 'Total number of ways'
     
  8. Aug 28, 2014 #7

    Orodruin

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    How many possibilities are there for who to give the bonus first? How many for the second? How many for the third? And so on. Remember that we are not going to give a bonus to someone who already got one.

    For the second question, consider that among the 3 who got $1000, it would have been equivalent if they would have gotten it in a different order. The same for the other amounts.
     
  9. Aug 28, 2014 #8
    The answer would be 15!


    15C7, 15C3, and 15C5
    am I correct?
     
  10. Aug 28, 2014 #9
    There are 7 out of 15 receiving $100 == There are 5 out of 15 receiving $3000== There are 3 out of 15 receiving $1000==.... (ABC=BCA=ACB=BAC=CAB=CBA)
    There are thus 6 ways.
     
  11. Aug 28, 2014 #10

    Orodruin

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    Exactly.

    Consider the group of 3. How many different ways can you order them? How many possibilities for the first person being chosen? How many for the second? How many for the third? (Note that we are now only concerned with the group of 3)
     
  12. Aug 28, 2014 #11
    That would be same as 'How many possibilities are there for who to give the bonus first?" right?
    so 3!
     
  13. Aug 28, 2014 #12
    $100: 15 ways for the 1st person, 14for the second....
    $1000:....
    $3000:...
    This is not correct. Because the question is how many possible ways to deliver 3 groups of (700,1000,3000) among 15 people. So 15 must too be divided into 3 groups (7,8,5). Since no order is mentioned. This is thus a permutation of 3 which is 3!=6.
     
    Last edited: Aug 28, 2014
  14. Aug 28, 2014 #13
    I am pretty sure it mean how many ways can the bonus be distributed to 15 people.
     
  15. Aug 28, 2014 #14
    Yes, I believe it is 6.
     
  16. Aug 28, 2014 #15

    Orodruin

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    No, this is a multinomial coefficient and it is certainly not 6. By your reasoning, there would be two different ways of splitting 545631 people in a group of 545600 and one of 31.
     
  17. Aug 28, 2014 #16

    Orodruin

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    Yes. And for the other groups?
     
  18. Aug 28, 2014 #17
    7!, 3! and 5!
     
  19. Aug 28, 2014 #18

    Orodruin

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    So, for each ordering of the entire set (of which there are 15!), there are how many equivalent orders? (Note that the equivalent orders within each group are independent)
     
  20. Aug 28, 2014 #19
    Yes, because you can't change the amount of bonus one will get in a company after the CEO has signed the awards.
    That John will get only $100 and Jess will receive $1000 is a fact. The way you split it will assign back Jess $100 and John $1000, which is incorrect and unfair and the company will go to court.
     
  21. Aug 28, 2014 #20
    I am not really sure if I get the question, so I would not try to answer it as any answer I give will be a guess.

    But I am assuming you meant "How many of the 15! are the same?" am I right.
     
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