Question

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?

Answer 1

do you know Combinations and Permutations

Answer 2

12C1 +12C2 +12C3+...12C8

Answer 3

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 .

Answer 4

are these your answers 50388 and 35846?

Answer 5

Let me run it through Wolfram Alpha, I just have the combinatoric form.

Answer 6

Unfortunately, no to the first one. I have 11C4, which is 165.
For the second, it was correct!

Answer 7

35846 is correct but 50388 is not?

Answer 8

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.

Answer 9

We just wrote that 8 bagels were determined, so the question boiled down to x1+x2...+x8=4

Answer 10

shouldn't you have

\[{8-1+12\choose 8-1}\] or \[{8-1+12\choose 12}\]

Answer 11

if you are using stars and bars

Answer 12

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.

Answer 13

you have 8 different bagels

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the bars make 8 possible locations or bagels

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Answer 14

let xxxxxxxxxxxx be the bagels

Answer 15

you want to arrange

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Answer 16

the above is an example of all 12 bagels of the same kind (the 8th kind of bagel)

Answer 17

the number of ways to arrange it is

\[{7+12\choose 7}={8-1+12\choose 8-1}\]

Answer 18

anyway...the number of ways to do problem 1 has to be larger than the number of was to do #2

Answer 19

since we are putting on a restriction

Answer 20

Ah! Okay!

Answer 21

You're correct, I wrote my response referring a different question. I agree with what you wrote!

Answer 22

so you know how to get the 35846

Answer 23

?

Answer 24

I know how to get the first answer.
For the second, would we use the inclusion - exclusion principle?

Answer 25

yes

Answer 26

All Solutions - solutions with x> 4 - solutions with y> 2 + intersection where (x> 4 and y> 2)?

How do I find all solutions?

Answer 27

that is it

we already did...\[{8-1+12\choose 8-1}=50388 \]

Answer 28

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.

Answer 29

no problem