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

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The discussion focuses on calculating the number of ways to select 10 jellybeans from red, blue, and green colors, with the condition of having at most 4 green jellybeans. The approach involves summing the combinations for selecting 1 to 4 green jellybeans, resulting in 10, 9, 8, and 7 ways respectively. There is also mention of the need to account for the case of selecting 0 green jellybeans, which was initially overlooked. The participants agree that the order of selection does not matter in this scenario. Overall, the total combinations are derived from these calculations, emphasizing the importance of including all cases.
Extreme112
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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.
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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