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

Revision as of 14:07, 22 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

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