Permutations

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thestag

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How many ways can 21 English & 19 Spanish books be arranged in a way that no two Spanish books are together?
 
I have tried to solve by finding P(19,22) But I don't think that's it. I mean I have no clue
 
Think about how you would arrange the books. Think about how you would arrange the books if you were to violate the given restraint.
 
thestag said:
How many ways can 21 English & 19 Spanish books be arranged in a way that no two Spanish books are together?
Click to expand...
This is a classic separator problem.. In some material on combinatorics it is know as the stars & bars problem.
Consider \(****||||||\), four stars and six bars. that string can be arranged in \(\dfrac{10!}{6!\cdot 4!}=210\) ways. Here is one way: \(*|~|~|**|~|*|\) We can see that the six bars create seven places \(\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_\) in which to place a star.
Thus there are \(\mathcal{C}_4^7=35\) ways to separate the stars.
If we have \(N~*'s\) and \(k~|'s\) then if \(k\ge N-1\) we can separate the stars using the bars in \(\mathcal{C}_N^{k+1}\) ways.
 
pka said:
This is a classic separator problem.. In some material on combinatorics it is know as the stars & bars problem.
Consider \(****||||||\), four stars and six bars. that string can be arranged in \(\dfrac{10!}{6!\cdot 4!}=210\) ways. Here is one way: \(*|~|~|**|~|*|\) We can see that the six bars create seven places \(\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_|\_\_\_\_\) in which to place a star.
Thus there are \(\mathcal{C}_4^7=35\) ways to separate the stars.
If we have \(N~*'s\) and \(k~|'s\) then if \(k\ge N-1\) we can separate the stars using the bars in \(\mathcal{C}_N^{k+1}\) ways.
Click to expand...

Thanks for your explanation. This is a whole new concept that I didn't know of. I think yt might be a good place to start.
 
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