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I am stuck on these two problems, can someone help?

 

There are 20 chairs in a row. Find the number of ways of choosing 5 of these chairs, so that no two chosen chairs are adjacent.

 

In how many ways can I distribute  6 identical cookies and 6 identical candies to 4 children, if each child must receive exactly 3 items?

 Dec 14, 2019
 #1
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First Problem: We can convert this to a problem where x1 + x2 + x3 + x4 + x5 + x6 = 18, where the xi are positive integers.  By stars and bars, the number of solutions is C(17,6) = 12376.

 

Second Problem: Let a, b, c, d be the number of cookies that the children get.  Then a + b + c + d = 6.  By stars and bars, the number of solutions is C(9,3) = 84.  But we must subtract the solutions where any of a, b, c, d is at least 4.  The number of such solutions is 12, so the answer is 84 - 12 = 72.

 Dec 14, 2019
 #2
avatar+108626 
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There are 20 chairs in a row. Find the number of ways of choosing 5 of these chairs, so that no two chosen chairs are adjacent.

 

Just for a starting reference.  If I hoose any 5 chairs from 20 then the answer would be 20C5= 15504 ways

so any answer must be less than that. 

 

Guests answer of 12376 passes that particular reasonable check but I would like him/her to expain much better how the answer was come by.

 

------------------------------------------------------

 

Here is my shot at this question.

 

Y stands for yes chosen

N stands for No not chosesn

*YN*YN*YN*YN*Y*

I interprete this as

Could be some chairs, yes gone, not gone, could be some chairs, etc

 

In order for 5 chairs to be yes chosen there must be at least  9 chairs altogether.

So it is the position of the other 11 chairs that is in question.    

The possible position of the other 11 chairs are represented by the stars.  There are 6 stars.

 

 

So my problem appears to have become.  How many ways can I put split 11 chairs into 6 groups.

11 stars and 5 bars  

11C5= 462 ways

 

I get 462 ways but it is quite likely wrong.   My error, if there is one, could be trivial or it could be a complete failure in logic.

 

 

Rollingblade, do you have an answer of any kind?

 Dec 14, 2019
edited by Melody  Dec 14, 2019
 #3
avatar+108626 
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Attn:  Rollingblade

 

I have just looked back at your more recent questions.

You have not responded to your answerers at all.

You used to respond but you seem to have stopped.

 

Please start again.

 

I and others put a lot of thought an effort into trying to help you. 

Even when our answers are not correct we still deserve a polite response.

 

And interaction is a key ingredient to learning.

 

Also
If you have any answer at all you should state what it is in the original question.

Why should we start out with less info than you have.

 Dec 14, 2019
edited by Melody  Dec 14, 2019
edited by Melody  Dec 15, 2019
 #4
avatar+993 
+1

Hi Rollingblade, 

 

I think I finally worked it out. 

 

Since the cookies and candies are identical, only the number of cookies affect each distribution. Therefore, the ways to distribute the cookies are \(a + b + c + d = 6\), where  \(0\le a,b,c,d\le 3\).

 

This is equal to 44, so the total ways to distribute is 44. 

 

For the chairs, there are 15 remaining chairs after 5 have been chosen. So \(a + b + c + d + e + f = 15\), where only a and f can be zero. The number of solutions are 4368. 

 

I hope this helped, 

Gavin

 Dec 15, 2019

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