(New page: By listing the possibilities, we get: 3 + 1 + 1 = 5 2 + 2 + 1 = 5 4 + 1 + 0 = 5 5 + 0 + 0 = 5 3 + 2 + 0 = 5 There are 2 ways if we want all the boxes to have an object, if not there are...)
 
 
(One intermediate revision by one other user not shown)
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3 + 1 + 1 = 5
 
3 + 1 + 1 = 5
 +
 
2 + 2 + 1 = 5
 
2 + 2 + 1 = 5
 +
  
 
4 + 1 + 0 = 5
 
4 + 1 + 0 = 5
 +
 
5 + 0 + 0 = 5
 
5 + 0 + 0 = 5
 +
 
3 + 2 + 0 = 5
 
3 + 2 + 0 = 5
  
 
There are 2 ways if we want all the boxes to have an object, if not there are 2 + 3 = 5 ways.
 
There are 2 ways if we want all the boxes to have an object, if not there are 2 + 3 = 5 ways.
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 +
Grading Comment:
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I believe the answer is correct. I did the same thing too. So, if I'm a grader for this course, I'll give you full mark. :)
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 +
--[[User:Mmohamad|hussaini]] 18:20, 5 October 2008 (UTC)

Latest revision as of 13:20, 5 October 2008

By listing the possibilities, we get:

3 + 1 + 1 = 5

2 + 2 + 1 = 5


4 + 1 + 0 = 5

5 + 0 + 0 = 5

3 + 2 + 0 = 5

There are 2 ways if we want all the boxes to have an object, if not there are 2 + 3 = 5 ways.

Grading Comment:

I believe the answer is correct. I did the same thing too. So, if I'm a grader for this course, I'll give you full mark. :)

--hussaini 18:20, 5 October 2008 (UTC)

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Questions/answers with a recent ECE grad

Ryne Rayburn