#Permutations and Combinations

17 messages · Page 1 of 1 (latest)

wild oak
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Hey Can I have some help with this question and the logic behind solving it..various methods of solving this would be greatly appreciated.

cosmic fossilBOT
fiery tundra
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So im not so good in combinatorics, but this sounds like a permutation to me

wild oak
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ok cool

fiery tundra
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So we know from the quest that our exact total marbles are 12 marbles, independet from the colours. The most important by questions like that is the acutual understanding of the question, so the part with the "?". Once you understand that 100% other things will mostly be more understandable for you.

Back to the solving part,

Total marbles: 12
Green marbles = 4
Blue marbles = 5
Red marbles = 3

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so every of these colours are identcal, thats why mostly permutations are liked to be used by these quests

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the first thing you can do is take a list of all combinations, but in the most cases that just takes too long. So i would go directly for the formular

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if you read ,,identical" then you can mostly assume that its a permutation quest

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because we do not have a exact order

mint dust
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If we have no limit to the number of marbles, you have three colors. You have three choices for the first marble, three choices for the second marble, three choices for the third marble, and three choices for the fourth marble. That's a total of 3^4 possible arrangements.

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Which possibilities does having only three red marbles eliminate?

mint dust
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Like red, green, red, green is distinct from red, green, green, red.

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The part about "the marbles of any given colour are identical" is to say that there's not red marble 3, green marble 2, red marble 1, green marble 4 as a distinct sequence from red marble 3, green marble 2, red marble 2, green marble 4.

wild oak
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Thank you I kinda undertsand this is there another way of thinking about it or solving it

mint dust
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You can list all possibilities and then cross off the ones that can't happen, then count the remaining possibilities.

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You can make a decision tree for that as well.