sh3lsh
  • sh3lsh
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
do you know Combinations and Permutations
anonymous
  • anonymous
12C1 +12C2 +12C3+...12C8
sh3lsh
  • sh3lsh
Oh, 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 .

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Zarkon
  • Zarkon
are these your answers 50388 and 35846?
sh3lsh
  • sh3lsh
Let me run it through Wolfram Alpha, I just have the combinatoric form.
sh3lsh
  • sh3lsh
Unfortunately, no to the first one. I have 11C4, which is 165. For the second, it was correct!
Zarkon
  • Zarkon
35846 is correct but 50388 is not?
sh3lsh
  • sh3lsh
Honestly, 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
  • sh3lsh
We just wrote that 8 bagels were determined, so the question boiled down to x1+x2...+x8=4
Zarkon
  • Zarkon
shouldn't you have \[{8-1+12\choose 8-1}\] or \[{8-1+12\choose 12}\]
Zarkon
  • Zarkon
if you are using stars and bars
sh3lsh
  • sh3lsh
My 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
  • Zarkon
you have 8 different bagels ||||||| the bars make 8 possible locations or bagels 1||||||| |2|||||| ||3||||| |||4|||| ||||5||| |||||6|| ||||||7| |||||||8
Zarkon
  • Zarkon
let xxxxxxxxxxxx be the bagels
Zarkon
  • Zarkon
you want to arrange |||||||xxxxxxxxxxxx
Zarkon
  • Zarkon
the above is an example of all 12 bagels of the same kind (the 8th kind of bagel)
Zarkon
  • Zarkon
the number of ways to arrange it is \[{7+12\choose 7}={8-1+12\choose 8-1}\]
Zarkon
  • Zarkon
anyway...the number of ways to do problem 1 has to be larger than the number of was to do #2
Zarkon
  • Zarkon
since we are putting on a restriction
sh3lsh
  • sh3lsh
Ah! Okay!
sh3lsh
  • sh3lsh
You're correct, I wrote my response referring a different question. I agree with what you wrote!
Zarkon
  • Zarkon
so you know how to get the 35846
Zarkon
  • Zarkon
?
sh3lsh
  • sh3lsh
I know how to get the first answer. For the second, would we use the inclusion - exclusion principle?
Zarkon
  • Zarkon
yes
sh3lsh
  • sh3lsh
All Solutions - solutions with x> 4 - solutions with y> 2 + intersection where (x> 4 and y> 2)? How do I find all solutions?
Zarkon
  • Zarkon
that is it we already did...\[{8-1+12\choose 8-1}=50388 \]
sh3lsh
  • sh3lsh
That's obvious. I don't even know why I asked that question. I got it from here , you're the bomb. Thanks so much.
Zarkon
  • Zarkon
no problem

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