# Homework Help: Combination is not my strong point

1. Aug 27, 2014

### SUXinPHY

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. Aug 27, 2014 ### Ray Vickson 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?

3. Aug 27, 2014

### SUXinPHY

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? 4. Aug 27, 2014 ### SUXinPHY 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. 5. Aug 28, 2014 ### Orodruin Staff Emeritus 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? 6. Aug 28, 2014 ### SUXinPHY 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' 7. Aug 28, 2014 ### Orodruin Staff Emeritus 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.

8. Aug 28, 2014

### SUXinPHY

The answer would be 15!

15C7, 15C3, and 15C5
am I correct?

9. Aug 28, 2014

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. 10. Aug 28, 2014 ### Orodruin Staff Emeritus 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) 11. Aug 28, 2014 ### SUXinPHY That would be same as 'How many possibilities are there for who to give the bonus first?" right? so 3! 12. Aug 28, 2014 ### Medicol$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
13. Aug 28, 2014

### SUXinPHY

I am pretty sure it mean how many ways can the bonus be distributed to 15 people.

14. Aug 28, 2014

### Medicol

Yes, I believe it is 6.

15. Aug 28, 2014

### Orodruin

Staff Emeritus
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.

16. Aug 28, 2014

### Orodruin

Staff Emeritus
Yes. And for the other groups?

17. Aug 28, 2014

### SUXinPHY

7!, 3! and 5!

18. Aug 28, 2014

### Orodruin

Staff Emeritus

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)

19. Aug 28, 2014

### Medicol

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.

20. Aug 28, 2014

### SUXinPHY

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.

21. Aug 28, 2014

### Orodruin

Staff Emeritus
No, this is not what I am doing. I am counting the number of possible ways of ordering the people. Then, for a given order, I am computing the number of equivalent orders which will give the same money distribution.

I am sorry to say, but your way of thinking about this is just plain wrong. By your argumentation, there should be 2 ways of distributing two bonuses of $1 and one bonus of$1000 to three people. So, let us take Alice, Bob, and Charlie. According to you, I need only consider two possible distributions so:
Distribution 1: Alice $1000, Bob$1, Charlie $1. Distribution 2: Alice$1, Bob $1000, Charlie$1.
If anyone is going to sue the company, it is Charlie.

Yes. Dividing the 15! with this number will give you the number of different distributions.

22. Aug 28, 2014

### SUXinPHY

Thanks a lot (I am assuming this would be the answer to the question). I will try to determine that number will get back to you soon.

23. Aug 28, 2014

### jz92wjaz

Does your teacher expect C notation, or factorial notation?

Either way, I recommend that you use the equation you provided in your original post and fill in the numbers for C for your original attempted answer, and then compare it with your new answer that you got by looking at it a different way.

24. Sep 2, 2014

### SUXinPHY

Thanks for helping just wanted to confirm the solution would be:
15!/ (3! * 7! * 5!)

Right?

25. Sep 2, 2014

### Orodruin

Staff Emeritus
Yes. This and the binomial coefficients are examples of the multinomial coefficients. As jz said, I also recommend that you write out the expression from your first post in terms of factorials and confirm that it is the same.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted