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August 20th, 2009, 07:48 PM  #1 
Newbie Joined: Aug 2009 Posts: 1 Thanks: 0  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+41)_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 
August 25th, 2009, 06:11 AM  #2 
Newbie Joined: Aug 2009 Posts: 2 Thanks: 0  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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