Final answer:
To find the number of arrangements for six math books and three English books on a shelf with books of each subject kept together, consider each subject as one unit and calculate permutations for each and multiple units. There are 2! ways to arrange the two units of books and 6! and 3! ways to arrange the books within each unit respectively, resulting in 8640 different arrangements.
Step-by-step explanation:
The problem you are asking about involves determining the number of different ways books can be arranged on a shelf under certain constraints. Specifically, you are asked to find how many ways six math books and three English books can be arranged if the math books are kept together and the English books are kept together. This is a combinatorial problem that can be solved using the principles of permutations and combinations.
To approach this problem, consider the six math books as a single unit and the three English books as another unit since they need to stay together. There are 2! or 2 different ways to arrange these two units on the shelf - either the math books come first or the English books come first. Within each unit, the books can be arranged among themselves. There are 6! ways to arrange the six math books and 3! ways to arrange the three English books.
Therefore, the total number of arrangements is the product of these permutations:
- 2! for the two units;
- 6! for the math books within their unit;
- 3! for the English books within their unit.
The calculation then becomes 2! * 6! * 3! = 2 * (6*5*4*3*2*1) * (3*2*1) = 2 * 720 * 6 = 8640 ways.