SOLUTION: 5 mathematics books, 5 physics books, and 5 chemistry books. In how many ways can they be arranged such that no two books of the same type are adjacent in the following cases: 1

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Question 1208370: 5 mathematics books, 5 physics books, and 5 chemistry books. In how many ways can they be arranged such that no two books of the same type are adjacent in the following cases:
1- In a straight line.
2- Around a circular table."
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
**1. In a straight line:**
We can think of this problem as arranging 15 books in a line with the constraint that no two books of the same type are adjacent.
**Step 1: Arrange the books of one type:**
* We can arrange 5 books of one type (say, math) in 5! ways.
**Step 2: Create gaps between the arranged books:**
* There are 6 gaps created between the 5 books. We need to place the remaining 10 books (5 physics and 5 chemistry) in these gaps.
**Step 3: Arrange the remaining books:**
* We can arrange 10 books in 10! ways.
However, we need to consider that the 10 books are not all distinct. We have 5 physics and 5 chemistry books. So, we need to divide by 5! for each type to account for identical permutations.
**Total arrangements:**
```
(5! * 10!) / (5! * 5!) = 10! / 5! = 30,240
```
**2. Around a circular table:**
For circular arrangements, we need to consider that any rotation of a specific arrangement is considered the same.
**Step 1: Arrange the books of one type in a line:**
* Arrange 5 books in a line: 5! ways.
**Step 2: Create gaps:**
* There are 5 gaps between the 5 books. Place the remaining 10 books in these gaps: 10! ways.
**Step 3: Account for rotations and identical permutations:**
* For circular arrangements, we divide by the number of positions (5 in this case) to account for rotations.
* We also divide by 5! for each type of book to account for identical permutations.
**Total arrangements:**
```
(5! * 10!) / (5 * 5! * 5!) = 10! / (5 * 5!) = 7,560
```
Answer by ikleyn(52754)   (Show Source): You can put this solution on YOUR website!
.
5 mathematics books, 5 physics books, and 5 chemistry books. In how many ways can they be arranged
such that no two books of the same type are adjacent in the following cases:
1- In a straight line.
2- Around a circular table."
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution to this problem in the post by @CPhill is partly incomplete
        and partly incorrect for both linear arrangements and circular arrangements.


    (1)   Why the solution by @CPhill is incomplete and incorrect for linear arrangements ? 

We can place 5 Math books in 5! = 120 ways.


Then we can make 4 gaps between these 5 Math books;
in addition to it, we have one "gap" before the row of 5 Math books and 
                           one "gap" after the row of 5 Math books.


Using these 4+1+1 = 6 gaps, we can place remaining books in pairs (Phys,Chem) 
in five leftmost gaps, or in 5 rightmost gaps.

    It gives the factor 2 based on placements.


In addition, we can permute Physics book in 5! ways and Chemistry books in 5! ways.


Thus, so far we have 2*5!*5!*5! ways.


But, in addition to it, we can permute (Phys,Chem) books inside each pair.

It gives   =  =  = 110592000 different possible arrangements.


So, even considering the same scheme of placement the books as @CPhill considered in his post, 
we see, that his counting is wrong.


In addition to this scheme, there are other placements, when some gaps between Math books may contain 
3 books like (Phys,Chem,Phys) or (Chem,Phys,Chem), while other gaps may contain only one book Phys or Chem,
It gives the number of other possible placements, missed by @CPhill, which I even do not try to count.


    (2)   Why the solution by @CPhill is incomplete and incorrect for circular arrangements ? 

Again, for circular arrangements, we can place Math books in   = 24 ways.

In circular arrangements, we can make then 5 gaps between five Math books.


In these 5 gaps, we can place 5 pairs (Phys,Chem) books.


We can permute Phys books in 5! ways and Chem books in 5! ways;
in addition, we can permute in 2 ways inside each pair.


It gives us 24*2^5*5!*5! =  = 691200 different circular arrangements.


    In addition to this scheme, there are other placements, when some gaps between Math books may contain 
    3 books like (Phys,Chem,Phys) or (Chem,Phys,Chem), while other gaps may contain only one book Phys or Chem,
    It gives the number of other possible placements, missed by @CPhill, which I even do not try to count.



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