A(3，3)*A(4，4)*A(5，5)*A(3，3)= 6 * 24 * 120 * 6 = 103680
2. If only math books must be placed next to each other (English books and Chinese books don't matter), how many permutations and combinations can there be?
(1) sorting out math book a (3,3),
(2) The math book is regarded as a book, and it is regarded as 10 book with other books, and the whole behavior is A( 10, 10).
So a (3,3) * a 3)*A( 10/0, 10)
3. If every two English books can't be placed next to each other, how many arrangements can there be?
Interpolation solutions of nonadjacent problems;
(1) The other eight books are arranged first: A (8,8)
(2) Insert five English books into the nine spaces generated by (1), one space and one A (9,5).
So a (8,8) * a (9,5)