Lets denote

the 10 novels be the elements of the set X = {x1, x2, x3, x4, x5, x6, x7, x8, x9, x10}

the 12 nonfiction be the elements of the set Y = {y1, y2, y3, y4, y5, y6, y7, y8, y9, y10, y11, y12}

the 4 plays be the elements of the set Z = {z1, z2, z3, z4}

The English teacher needs to pick 11 books: 4 novels, 4 nonfiction and 3 plays.

Lets assume that the order of the books is not important that means that we deal with combinations.

A combination is a way of selecting several things out of a larger group, where order does not matter.

The number of k-combinations from a given set S of n elements is often denoted by C(n, k) so in our case by using the formula for combinations without repetitions C(n,k) = n! / (k!*(n - k!) we have

- the number of combinations of 4 objects from the elements of the set X of 10 elements

C(4,10) = 10! / (4! * (10 - 4)!) = 10! / (4! 6!) = 7*8*9*10/ 2*3*4 = 210

- the number of combinations of 4 objects from the elements of the set Y of 12 elements

C(4,12) = 12! / (4! * (12 - 4)!) = 12! / (4! 8!) = 9*10*11*12 / 2*3*4 = 494

- the number of combinations of 3 objects from the elements of the set Z of 9 elements

C(3,9) = 9! / (3! * (9 - 3)!) = 10! / (3! 6!) = 7*8*9*10/ 2*3 = 840

The number we are looking for is the product

C(4,10) * C(4,12) * C(3,9) = 210 * 494 * 840 = 87,141,600

The English teacher can put the books in 87,141,600 different ways.