How Many Ways to Select 10 Jellybeans With Up to 4 Greens?

Click For Summary

Homework Help Overview

The problem involves selecting 10 jellybeans from three colors: Red, Blue, and Green, with the constraint that no more than 4 of the jellybeans can be Green. Participants are exploring the combinatorial aspects of this selection process.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss various ways to count the selections based on the number of Green jellybeans chosen, considering different cases (0 to 4 Greens). There is an exploration of how the order of selection affects the counting, with some questioning the completeness of their counts.

Discussion Status

Some participants have provided partial counts for different cases of Green jellybeans, while others are questioning assumptions about the order of selection and whether all cases have been accounted for. There is acknowledgment of a potential oversight regarding the case of selecting 0 Green jellybeans.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The discussion reflects a collaborative effort to clarify the problem and ensure all scenarios are considered.

Extreme112
Messages
5
Reaction score
0

Homework Statement


How many ways can you select 10 jellybeans from colors Red, Blue, Green so that at most you only have 4 Green jellybeans?

Homework Equations


...

3. The Attempt at a Solution [/B]
# of ways = # of ways to pick 1 Green + # of ways to pick 2 Green + #of ways to pick 3 Green + # of ways to pick 4 Green.

1 Green jellybean: After picking out the jellybean, there are then 9 left to choose from.
* * * * $ * * * * *
If the '*' are the 9 jellybeans and '$' is the divider to separate the jellybeans so that those on the left of it are Red and those to the right of it are Blue then there are 10!/9! or 10 ways to rearrange it.

2 Greens: Following the same process above would result in 9!/8! = 9
3 Greens: 8!/7! = 8
4 Greens: 7!/6! = 7

Therefore you would have 10+9+8+7 ways to select 10 jellybeans with at most having 4 Green jellybeans.
 
Physics news on Phys.org
For two greens,
one comes out first, the second one can come out with the second draw, or the third, or... up to the tenth.
that's nine ways... but the first could have come out on the 3rd draw, with the second coming out on the 4th or subsequent... that's another 6 ways or something isn't it?
So that's 15 ways to get 2 green, and I haven't finished counting yet.
 
Simon Bridge said:
For two greens,
one comes out first, the second one can come out with the second draw, or the third, or... up to the tenth.
that's nine ways... but the first could have come out on the 3rd draw, with the second coming out on the 4th or subsequent... that's another 6 ways or something isn't it?
So that's 15 ways to get 2 green, and I haven't finished counting yet.
I would say Extreme112 is interpreting the question correctly, that the order of selection is unimportant.
Extreme112 said:
you would have 10+9+8+7 ways to select 10 jellybeans with at most having 4 Green jellybeans.
You left out one case.
 
haruspex said:
I would say Extreme112 is interpreting the question correctly, that the order of selection is unimportant.

You left out one case.
I think I forgot the 0 case. Thanks for the help guys.
 

Similar threads

Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
1K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K