Question

A bookseller has seven different textbooks on solid mechanics, five different texts on fluid mechanics, and four on thermodynamics. She wants to make a window display consisting of three books on each subject, arranged in such a way that books on the same topic are together. How many displays are possible?

Answer 1

There are 7P3 permutations possible of the books on solid mechanics, 4P3 permutations possible of the books on fluid mechanics and 4P3 permutations possible of the books on thermodynamics. Each permutation can be taken with each of the other possible permutations. Also there are 3! permutations possible of the three groups of books.
Therefore the number of possible displays is given by
\[\frac{7!}{4!}\times \frac{5!}{2!}\times 4!\times 3!=you\ can\ calculate\]

Answer 2

Thank you! :D