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

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

do you know Combinations and Permutations

- anonymous

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

- 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 .

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Zarkon

are these your answers 50388 and 35846?

- sh3lsh

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

- sh3lsh

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

- Zarkon

35846 is correct but 50388 is not?

- 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

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

- Zarkon

shouldn't you have
\[{8-1+12\choose 8-1}\] or \[{8-1+12\choose 12}\]

- Zarkon

if you are using stars and bars

- 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

you have 8 different bagels
|||||||
the bars make 8 possible locations or bagels
1|||||||
|2||||||
||3|||||
|||4||||
||||5|||
|||||6||
||||||7|
|||||||8

- Zarkon

let xxxxxxxxxxxx be the bagels

- Zarkon

you want to arrange
|||||||xxxxxxxxxxxx

- Zarkon

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

- Zarkon

the number of ways to arrange it is
\[{7+12\choose 7}={8-1+12\choose 8-1}\]

- Zarkon

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

- Zarkon

since we are putting on a restriction

- sh3lsh

Ah! Okay!

- sh3lsh

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

- Zarkon

so you know how to get the 35846

- Zarkon

?

- sh3lsh

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

- Zarkon

yes

- 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

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

- 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

no problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.