Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > University Math Help > Discrete Mathematics, Set Theory and Logic
Reply
 
Thread Tools Display Modes
  #1  
Old November 6th, 2009, 02:16 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Question Counting

A candy factory has an endless supply of red, orange, yellow, green, blue, and violet jelly beans. The factory packages the jelly beans into jars of 84 jelly beans each. As a marketing gimmick, the factory guarantees that no two jars have exactly the same number of red, orange, yellow, green, blue, and violet jelly beans. What is the maximum number of jars the factory can produce?
Reply With Quote
Advertisement
 
  #2  
Old November 6th, 2009, 02:26 PM
MHF Contributor

 
Join Date: Aug 2006
Posts: 6,715
Thanks: 68
Thanked 2,485 Times in 2,279 Posts
Plato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond repute
Default

Quote:
Originally Posted by essedra View Post
A candy factory has an endless supply of red, orange, yellow, green, blue, and violet jelly beans. The factory packages the jelly beans into jars of 84 jelly beans each. As a marketing gimmick, the factory guarantees that no two jars have exactly the same number of red, orange, yellow, green, blue, and violet jelly beans. What is the maximum number of jars the factory can produce?
How many non-negative integral solutions are there to this equation are there:
r+o+y+g+b+v=84~?
Reply With Quote
  #3  
Old November 6th, 2009, 02:31 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Default

Quote:
Originally Posted by Plato View Post
How many non-negative integral solutions are there to this equation are there:
r+o+y+g+b+v=84~?
I didn't get what you mean...
Reply With Quote
  #4  
Old November 6th, 2009, 02:33 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Default

I tried to solve it like this:

6x6x6x6x6x6x6x6x6x6x6x6x6x6x6........ (84 times) =

6^84 = 2 + 65 zeros

which is same as

2x10^65

but the answer was incorrect...
Reply With Quote
  #5  
Old November 6th, 2009, 02:42 PM
MHF Contributor

 
Join Date: Aug 2006
Posts: 6,715
Thanks: 68
Thanked 2,485 Times in 2,279 Posts
Plato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond repute
Default

Quote:
Originally Posted by essedra View Post
I didn't get what you mean...
Somewhere in you text-material there must be a section of making multi-selections.
How many ways to put N identical objects into K different cells: \binom{N+K-1}{N}.

So when I asked about many solutions to r+o+y+g+b+v=84,
that is equivalent putting 84 identical ones into six different colors.
Each solution is one way the candy factory can package the 84 jelly beans.
Reply With Quote
  #6  
Old November 6th, 2009, 02:53 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Default

I still couldn't solve the problem, because i'm confused about the "k" you're talking about...are order and repetition important?
Reply With Quote
  #7  
Old November 6th, 2009, 03:02 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Default

I would be really grateful if you could show me the way to solve this problem...
Reply With Quote
  #8  
Old November 6th, 2009, 03:08 PM
MHF Contributor

 
Join Date: Aug 2006
Posts: 6,715
Thanks: 68
Thanked 2,485 Times in 2,279 Posts
Plato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond repute
Default

Quote:
Originally Posted by essedra View Post
I still couldn't solve the problem, because i'm confused about the "k" you're talking about...are order and repetition important?
Let me give you an example.
Suppose that an ice cream shop has 21 flavors.
It runs a special offer of a plate of 4 scoops of any combinations.
How many different offerings are there?
Answer: \binom{4+21-1}{4}=\frac{24!}{(4!)(20!)}

That is the number of ways to put 4 identical objects (scoops) into 21 (flavors) different cells.
Note that the 4 can be all different, all the same, or say two of one and two different.
Reply With Quote
  #9  
Old November 6th, 2009, 03:21 PM
Junior Member
 
Join Date: Mar 2009
Posts: 33
Thanks: 13
Thanked 0 Times in 0 Posts
essedra is on a distinguished road
Default

so I applied what you said...

89!/(6!*84! ) = 6917940,3333333333333333333333333

but the result was incorrect...
Reply With Quote
  #10  
Old November 6th, 2009, 03:26 PM
MHF Contributor

 
Join Date: Aug 2006
Posts: 6,715
Thanks: 68
Thanked 2,485 Times in 2,279 Posts
Plato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond reputePlato has a reputation beyond repute
Default

Quote:
Originally Posted by essedra View Post
so I applied what you said...

89!/(6!*84! ) = 6917940,3333333333333333333333333

but the result was incorrect...
I don't think that you have caculated that correctly.
\binom{84+6-1}{84}=\binom{89}{84}=41507642
Reply With Quote
The following users thank Plato for this useful post:
Donate to MHF
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 12:55 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.