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I'm not sure if I'm thinking about this the right way, but here is what I came up with. There are <math>{15 \choose 1}=15</math> ways to put an object in box 1. With 14 objects left, there are <math>{14 \choose 2}</math> ways to put 2 objects in box 2. This continues until you have <math>{5 \choose 5}=1</math> ways to put 5 objects in box 5. Multiplying these terms together, the answer is 37,837,800, which I thought seemed high even for this kind of problem. Did anyone else get this? | I'm not sure if I'm thinking about this the right way, but here is what I came up with. There are <math>{15 \choose 1}=15</math> ways to put an object in box 1. With 14 objects left, there are <math>{14 \choose 2}</math> ways to put 2 objects in box 2. This continues until you have <math>{5 \choose 5}=1</math> ways to put 5 objects in box 5. Multiplying these terms together, the answer is 37,837,800, which I thought seemed high even for this kind of problem. Did anyone else get this? | ||
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I was in the study group when we got this same answer. I believe its right! --[[User:Hschonho|Mike Schonhoff]] 16:20, 24 September 2008 (UTC) | I was in the study group when we got this same answer. I believe its right! --[[User:Hschonho|Mike Schonhoff]] 16:20, 24 September 2008 (UTC) | ||
Latest revision as of 12:20, 24 September 2008
I'm not sure if I'm thinking about this the right way, but here is what I came up with. There are $ {15 \choose 1}=15 $ ways to put an object in box 1. With 14 objects left, there are $ {14 \choose 2} $ ways to put 2 objects in box 2. This continues until you have $ {5 \choose 5}=1 $ ways to put 5 objects in box 5. Multiplying these terms together, the answer is 37,837,800, which I thought seemed high even for this kind of problem. Did anyone else get this?
I was in the study group when we got this same answer. I believe its right! --Mike Schonhoff 16:20, 24 September 2008 (UTC)
