- 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)
- Stars and bars is a good way to look at this problem, and while the math behind it will work out to equal the same, perhaps it would make more since to remember we are choosing 12 croisants (aka the stars) not the bars, so it should be (12 + 6 - 1) choose 12. Again, this is the same as (12 + 6 - 1) choose 5 by the reasoning that you are counting the same thing just in a different way, but for the concept of the question, I feel is is more correct to be choosing 12.--mnoah 16:17, 24 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!!
