# Counting

• Nov 6th 2009, 12:16 PM
essedra
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?
• Nov 6th 2009, 12:26 PM
Plato
Quote:

Originally Posted by essedra
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:
$\displaystyle r+o+y+g+b+v=84~?$
• Nov 6th 2009, 12:31 PM
essedra
Quote:

Originally Posted by Plato
How many non-negative integral solutions are there to this equation are there:
$\displaystyle r+o+y+g+b+v=84~?$

I didn't get what you mean...
• Nov 6th 2009, 12:33 PM
essedra
I tried to solve it like this:

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

6^84 = 2 + 65 zeros

which is same as

2x10^65

• Nov 6th 2009, 12:42 PM
Plato
Quote:

Originally Posted by essedra
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: $\displaystyle \binom{N+K-1}{N}$.

So when I asked about many solutions to $\displaystyle r+o+y+g+b+v=84$,
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.
• Nov 6th 2009, 12:53 PM
essedra
I still couldn't solve the problem, because i'm confused about the "k" you're talking about...are order and repetition important?
• Nov 6th 2009, 01:02 PM
essedra
I would be really grateful if you could show me the way to solve this problem...
• Nov 6th 2009, 01:08 PM
Plato
Quote:

Originally Posted by essedra
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: $\displaystyle \binom{4+21-1}{4}=\frac{24!}{(4!)(20!)}$

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.
• Nov 6th 2009, 01:21 PM
essedra
so I applied what you said...

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

but the result was incorrect...
• Nov 6th 2009, 01:26 PM
Plato
Quote:

Originally Posted by essedra
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.
$\displaystyle \binom{84+6-1}{84}=\binom{89}{84}=41507642$