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sh3lsh
 one year ago
Stars 'n' bars Theorem Problem.
1. A bagel shop has 8 kinds of bagels. How many ways to buy a dozen bagels?
2. A bagel shop has 8 kinds of bagels. How many ways to buy a dozen bagels, with at most 4 onion and at most 2 poppy seed?
sh3lsh
 one year ago
Stars 'n' bars Theorem Problem. 1. A bagel shop has 8 kinds of bagels. How many ways to buy a dozen bagels? 2. A bagel shop has 8 kinds of bagels. How many ways to buy a dozen bagels, with at most 4 onion and at most 2 poppy seed?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know Combinations and Permutations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.012C1 +12C2 +12C3+...12C8

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Oh, are you sure? I know the answer for both problems, I'm just looking for the methods. The first solution should have k=4;n=8 .

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1are these your answers 50388 and 35846?

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Let me run it through Wolfram Alpha, I just have the combinatoric form.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately, no to the first one. I have 11C4, which is 165. For the second, it was correct!

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.135846 is correct but 50388 is not?

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0Honestly, perhaps I'm concluding the wrong answer for the first question, we didn't complete the problem to its entirety. I just have k =4; n=8, which in the stars n bars theorem is just (10 choose 8). Perhaps it meant something else.

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0We just wrote that 8 bagels were determined, so the question boiled down to x1+x2...+x8=4

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1shouldn't you have \[{81+12\choose 81}\] or \[{81+12\choose 12}\]

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1if you are using stars and bars

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0My understanding that it's a little different with a lower and upper bound. I don't understand the concepts yet to agree nor disagree with what you wrote.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you have 8 different bagels  the bars make 8 possible locations or bagels 1 2 3 4 5 6 7 8

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1let xxxxxxxxxxxx be the bagels

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you want to arrange xxxxxxxxxxxx

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1the above is an example of all 12 bagels of the same kind (the 8th kind of bagel)

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1the number of ways to arrange it is \[{7+12\choose 7}={81+12\choose 81}\]

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1anyway...the number of ways to do problem 1 has to be larger than the number of was to do #2

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1since we are putting on a restriction

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0You're correct, I wrote my response referring a different question. I agree with what you wrote!

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1so you know how to get the 35846

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0I know how to get the first answer. For the second, would we use the inclusion  exclusion principle?

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0All Solutions  solutions with x> 4  solutions with y> 2 + intersection where (x> 4 and y> 2)? How do I find all solutions?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1that is it we already did...\[{81+12\choose 81}=50388 \]

sh3lsh
 one year ago
Best ResponseYou've already chosen the best response.0That's obvious. I don't even know why I asked that question. I got it from here , you're the bomb. Thanks so much.
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