Question 1151727: A bag contains four red marbles, one green one, one lavender one, three yellows,
and two orange marbles.
How many sets of five marbles include at most one of the yellow ones?
Found 2 solutions by Edwin McCravy, jim_thompson5910:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! A bag contains four red marbles, one green one, one lavender one, three yellows,
and two orange marbles.
How many sets of five marbles include at most one of the yellow ones?
There are 4+1+1+1+3+2 = 12 marbles
Remove 2 of the 3 yellow marbles, then choose any 4 from the remaining 10 marbles,
3 yellow marbles CHOOSE 2 to remove in 3C2 = 3 ways.
TIMES
10 remaining marbles CHOOSE 5 = 10C5 = 252 ways.
Answer 3∙252 = 756 ways
Edwin
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
If there is a way to distinguish between the marbles of the same color, say there's a numeric label on each marble, then the answer is 756 as Edwin has shown.
If there isn't a way to tell two red marbles apart, and similar for the other colors, then the answer is 23.
Below shows the work on how to get 23.
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We have this multiset
{R, R, R, R, G, L, Y, Y, Y, A, A}
where
R = red marble
G = green marble
L = lavender marble
Y = yellow marble
A = orange marble (I'm using the letter "A" instead of the letter "O" because it is easilly mixed up with the number zero)
We can write this shorthand notation
4R, 1G, 1L, 3Y, 2A
to indicate 4 red, 1 green, 1 lavender, 3 yellow, 2 orange
We want to select five marbles such that we only have at most 1 yellow marble. This means we either have 1 yellow or no yellows. So we can toss out 2 of the yellow marbles getting this updated notation
4R, 1G, 1L, 1Y, 2A
We are able to toss any two yellow marbles because we can't tell them apart.
Now consider the number of ways we can add to 5 using positive integers. There are 6 ways to do this
4+1
3+2
3+1+1
2+2+1
2+1+1+1
1+1+1+1+1
These are considered partitions and will help us figure out how to select the marbles in an organized way.
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Partition 4+1
We have 4 red marbles. No other color has this many marbles. So the '4' applies to red only. The '1' applies to any of the other colors. We have 4 other colors, so we have 4 ways to fill out this partition
4R,1G
4R,1L
4R,1Y
4R,1A
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Partition 3+2
No color has 3 marbles other than red, so the '3' applies to the red marbles. No other color than orange has 2 marbles (recall we tossed out 2 yellows leaving with just 1), so the '2' applies to orange
There is only one way we can have 3 marbles of one color and 2 marbles of another color
3R, 2A
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Partition 3+1+1
The '3' applies to the red marbles. The other '1's are for the other colors
3R,1Y,1A
3R,1L,1A
3R,1L,1Y
3R,1G,1A
3R,1G,1Y
3R,1G,1L
There are 6 ways to fill this section out.
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Partition 2+2+1
The '2's apply to the red and orange marbles. The '1' is for the other colors
2R,1Y,2A
2R,1L,2A
2R,1G,2A
There are 3 ways to do it here.
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Partition 2+1+1+1
The '2' either applies to red marbles or orange marbles. The '1's are for anything else
2R,1L,1Y,1A
2R,1G,1Y,1A
2R,1G,1L,1A
2R,1G,1L,1Y
1G,1L,1Y,2A
1R,1L,1Y,2A
1R,1G,1Y,2A
1R,1G,1L,2A
There are 8 rows in the list above.
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Partition 1+1+1+1+1
There's only one way to do this selection and that is to simply pick one of each color
1R,1G,1L,1Y,1A
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Add up the subtotals:
4+1+6+3+8+1 = 23
Here are all 23 ways to select five marbles in which we cannot tell similar color ones apart (eg: we can't tell two red marbles apart)
| Number | Set |
| 1 | 4R,1G |
| 2 | 4R,1L |
| 3 | 4R,1Y |
| 4 | 4R,1A |
| 5 | 3R,2A |
| 6 | 3R,1Y,1A |
| 7 | 3R,1L,1A |
| 8 | 3R,1L,1Y |
| 9 | 3R,1G,1A |
| 10 | 3R,1G,1Y |
| 11 | 3R,1G,1L |
| 12 | 2R,1Y,2A |
| 13 | 2R,1L,2A |
| 14 | 2R,1G,2A |
| 15 | 2R,1L,1Y,1A |
| 16 | 2R,1G,1Y,1A |
| 17 | 2R,1G,1L,1A |
| 18 | 2R,1G,1L,1Y |
| 19 | 1G,1L,1Y,2A |
| 20 | 1R,1L,1Y,2A |
| 21 | 1R,1G,1Y,2A |
| 22 | 1R,1G,1L,2A |
| 23 | 1R,1G,1L,1Y,1A |
The order of the selection does not matter.
So 1R,1G,1L,1Y,1A is the same as 1G,1R,1L,1Y,1A
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