BarneyStinson wrote:
11 english books can be arranged in 11 slots (say) in 11! ways.
There will be 10 empty spaces between adjacent english books and another 2 empty spaces at either ends into which the french books can be placed such that no two french books are adjacent to one another. This is done in 12C9 ways.
Total number of ways should therefore be 12C9 * 11!.
Not so.
Let's say we have A1, A2, B1, B2 (meaning that A-s and B-s are distinct). We want to arrange them so that no B-s are adjacent:
*A1*A2* and we can place B1 and B2 in 3 empty slots. It can be done in 3C2 # of ways. BUT A1 and A2 can be arranged like *A1*A2* OR *A2*A1*, plus B1 and B2 can be arranged as B1B2 or B2B1.
Total # of ways 3C2*2!*2!=12.
Still if not convinced:
B1,A1,B2,A2
B1,A2,B2,A1
B2,A1,B1,A2
B2,A2,B1,A1
A1,B1,A2,B2
A2,B1,A1,B2
A1,B2,A2,B1
A2,B2,A1,B1,
B1,A1,A2,B2
B1,A2,A1,B2
B2,A1,A2,B1
B2,A2,A1,B1
BUT again this is the case when we have DISTINCT items. So, if we were told that all French book are different and all English books are different, then the answer would be 12C9*11!*9!.
In our original question we are not told that French books are different and are not told that English books are different. So # of ways would be 12C9.
Let's consider the easier example: # of ways to arrange two A-s and 2 B-s so that no B-s are adjacent: 3C2=3.
*A*A*
BABA
ABAB
BAAB
Hope it helps.