(New page: Would part A just be C(12,7)? Where 12 is how many croissants you can choose from and 7 is how many types of croissants there are to choose from? Wouldn't it be C(12,6) with repetition si...)
 
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Isn't this entire problem about the bars and stars? You have 6 types, 12 to choose. Use 5 bars and 12 stars: |***|***|**|**|** for A. So, the answer is (12 + 6 - 1) choose 5. And for C, you basically just select 2*6 from the stacks first. This leaves you with 12 more to select. So the answer to C is the answer to A.
 
Isn't this entire problem about the bars and stars? You have 6 types, 12 to choose. Use 5 bars and 12 stars: |***|***|**|**|** for A. So, the answer is (12 + 6 - 1) choose 5. And for C, you basically just select 2*6 from the stacks first. This leaves you with 12 more to select. So the answer to C is the answer to A.
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*Yea stars and bars!!!--[[User:Aifrank|Aifrank]] 18:22, 23 September 2008 (UTC)
  
 
I thought it was about the croissants! Am I in the wrong section?
 
I thought it was about the croissants! Am I in the wrong section?
  
 
I hope this helps, I don't know anything about technology, but I posted on this and wanted to have the link come up... hope it works this time!!
 
I hope this helps, I don't know anything about technology, but I posted on this and wanted to have the link come up... hope it works this time!!

Revision as of 14:22, 23 September 2008

Would part A just be C(12,7)? Where 12 is how many croissants you can choose from and 7 is how many types of croissants there are to choose from?

Wouldn't it be C(12,6) with repetition since there's 6 different type of croissants and after you choose one you can choose it again if you want to. Yeah nevermind, I mean't C(12,6).

For part C, how could you compensate for counting "at least 2 of each croissant"?

For part C and the following problems, I had a+b+c+d+e+f=24 and we know that a>=2 so then we have a'=a-2... So with the following we have (a'+2)+(b'+2)+(c'+2)...+(f'+2)=24. So then subtracting all of the 2s, we have a'+b'+c'+d'+e'+f'=12. So then we have C(12+6-1,6)= C(17,6)=12,376

Isn't this entire problem about the bars and stars? You have 6 types, 12 to choose. Use 5 bars and 12 stars: |***|***|**|**|** for A. So, the answer is (12 + 6 - 1) choose 5. And for C, you basically just select 2*6 from the stacks first. This leaves you with 12 more to select. So the answer to C is the answer to A.

  • Yea stars and bars!!!--Aifrank 18:22, 23 September 2008 (UTC)

I thought it was about the croissants! Am I in the wrong section?

I hope this helps, I don't know anything about technology, but I posted on this and wanted to have the link come up... hope it works this time!!

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett