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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 | ||
| + | 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 13:07, 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
