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August 20th, 2009, 07:48 PM   #1
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Number of choices...

I have indicated what I think is the answer but I think I may be wrong.

Given:

9 different cups
9 similar glasses
4 different tea bags
4 similar tea spoons

note - no problem with multiple items in a glass or cup

a) Spoons amongs glasses

(All spoons in 1 glass) + (2 spoons in some glass + 2 spoons in another glass) + (1 spoon in a glass + 3 spoons in another glass) = 3 choices

b) Spoons amongst cups

Pick 4 cups from 9 w/ repition: (9+4-1)_C_4 = 12C4 = 495

c) Tea bags amongst cups

9^4

d) Tea bags amongs glasses

12 -- a) part * 4 since tea bags are distinct
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August 25th, 2009, 06:11 AM   #2
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Re: Number of choices...

Well, I have to say my answer is somehow different from yours.
In the part of spoon with glass, we are the same.
In the next part, my answer as follows:

when all the spoons with only one cup, the number of choice is : 9

when two of them with one cup, the one of the rest with another cup, and the last spoon with another three cup(2+1+1), the number of the choice is : 9*8!/(2!*6!)=252

when two of them with on cup, the rest of the spoons with another cup(2+2), the answer is : 9*8=72

when three of the spoons with one cup, the last one with another cup, the answer is : 9*8=72

when each spoons with single cup, the answer is : 9!/(4!*5!)=126
so the answer of the problem about the spoons with cups is : 9+252+72+72+126=531

By the way, I am not quite clear about the meaning of "no problem with multiple items in a glass or cup"
Besides, I am not sure the bags and glasses problem. Could you please illustrate the problem more clear? bags with glasses or glasses with bags? hope my anwser will be useful
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