# Math Help - Jelly Bean Combination/Permutation Problem

1. ## Jelly Bean Combination/Permutation Problem

Fun questions, but I am having problems setting up the problem. I believe I got (a) by using C(300, 10). (b), (c), and (d), I am confuse on how to begin or what to think about.
I know order does not matter. Should I treat these problems as unordered selection of objects and use C(n, r) = n! / r! (n - r)! ?

Consider a bag of jelly beans that has 100 red, 100 yellow, and 100 green jelly beans.

a) How many color combinations can you get by drawing 10 beans from the bag?

b) How many color sequences can you get by drawing 5 beans from the bag?

c) How many color combinations of 10 beans have exactly two yellow beans?

d) How many color combinations of 10 beans have at least two yellow beans?

2. ## Re: Jelly Bean Combination/Permutation Problem

a)I think you're wrong. two sets of 10 elements may represent the same combination of colors.
b)it's easy, you have 3 possibilities for each element in the sequence, so 3^5.
c)each color combinations of 10 beans that have exactly two yellow beans can be represented by a color combination of 8 beans that have only red and green, so you count how many combinations of 8 red/green beans you can have.
d)think of the complement of a set. you use the result of question a) and remove the number of combinations of 10 beans that have 1 or 0 yellow beans.

ps : maybe i'm wrong, i'm not sure what they mean when they say a color combination, are two sets of different beans but with same colors considered as two different combinations ? (i supposed no)

3. ## Re: Jelly Bean Combination/Permutation Problem

Originally Posted by inflames098

Consider a bag of jelly beans that has 100 red, 100 yellow, and 100 green jelly beans.

a) How many color combinations can you get by drawing 10 beans from the bag?
b) How many color sequences can you get by drawing 5 beans from the bag?
c) How many color combinations of 10 beans have exactly two yellow beans?
d) How many color combinations of 10 beans have at least two yellow beans?
This is a very deceptive problem. It is not asking what you think it is. It is asking about color combinations.

The number of ways to put $N$ identical objects into $K$ distinct cells is $\binom{N+K-1}{N}$

In this case $N$ is the number of jellybeans chosen and $K$ is the number of different colors to chose from.

It is the number of whole number solutions to $R+Y+G=10$.

For part c) Take $Y=2$ and solve $R+G=8$ as we have above.