Maybe smart guys can see the close form solution right away. Here is one way tackling this problem step by step. Let's forget about the ,000 and assume there are $20 in total. Since the minumal investment is $1, we assign $1 to each project. Now we have $16 left. We need to assign $16 to 4 projects, with the possibility of assign $0 to some projects.
Denote this number by f($16,4), where 16 represents money and 4 represents the number of projects. We now can use dynamic programming.
It is easy to see that f($n,1)=0, f($n, 2)=n+1, and f($n, 3)=f($n,2)+f($n-1,2)+...+f($1,2)+f($0,2)=(n+1)+n+...+2+1=(n*n+3n+2)/2.
Thus, f($n,4)=f($n,3)+f($n-1,3)+...+f($0,3)=[(n*n+3n+2)+...+(1*1+3*1+2)+(0*0+3*0+2)]/2.
The answer to your first question is f($16,4)= HALF OF (16*16+...+1*1)+3*(16+...+1)+17*2.
The answer to your second question is f($16, 5), in which there is a dummy project absorbing the rest money. Of course, you need one more step of recursion.
Does anyone have a better solution procedure?
回复:请教解数学题。
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回复:回复:请教解数学题。
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09/07/2010 postreply
19:44:24
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回复:回复:回复:请教解数学题。
-guest1000-
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09/07/2010 postreply
19:51:53
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第二题答案是不是:C(16+3,3)+C(15+3,3)+...+C(1+3,3)+C(0+3,3)
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09/08/2010 postreply
09:10:55
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I think we should change 16 to 20....we get C(24,4)=...
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09/08/2010 postreply
09:35:30
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how to understand: must be invested among 4 possible opportuniti
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09/08/2010 postreply
10:00:33
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even if we should change 16 to 20....we get C(21,4)=...
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09/08/2010 postreply
10:03:46
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sorry, C(21,4)=... wrong; C(24,4)=... right,if we should change
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09/08/2010 postreply
10:07:18
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you are right:C(20,4)=C(16+3,3)+C(15+3,3)+...+C(1+3,3)+C(0+3,3)
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09/08/2010 postreply
10:09:50
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第二题:What if not all money need be invested? 可这样理解:
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09/08/2010 postreply
10:37:04
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我认为,您的第一题答案是正确的。
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09/08/2010 postreply
09:03:17
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回复:我认为,您的第一题答案是正确的。
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09/08/2010 postreply
12:31:24
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yes,您的第二题答案是正确的,as well。
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09/08/2010 postreply
13:03:14
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Put 16 balls in 5 bins, no matter 5 bins can be empty or not, wh
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09/08/2010 postreply
13:17:31