A math teacher uses four algebra books, two geometry books, and three pre-calculus books for reference. In how many ways can the teacher arrange the books on the shelf if books coiver the same subject matter are kept together?
There are 3! =6 ways to order the three separate topics.
For each of these ways, there are 4! ways to arrange the algebra books, 2! ways for the geometry books, and 3! ways for the pre-cal books.
So, in all, there are 3! 4! 2! 3! = 6*24*2*6 = 1728 ways.
Note that if the subjects did not have to be kept together, there would be 9! = 362880 ways.
To solve this problem, we need to consider the groups of books that cover the same subject matter as individual units. We have four groups of books:
- Algebra books (4 books)
- Geometry books (2 books)
- Pre-calculus books (3 books)
First, let's calculate the number of ways we can arrange each group individually.
1. For the algebra books, there are 4! (4 factorial) ways to arrange them.
2. For the geometry books, there are 2! (2 factorial) ways to arrange them.
3. For the pre-calculus books, there are 3! (3 factorial) ways to arrange them.
Now, let's calculate the total number of ways to arrange all the books together.
Since the groups of books cover the same subject matter and need to be kept together, we treat each group as a single unit. Therefore, we can arrange the three units (algebra, geometry, pre-calculus) in 3! ways.
Hence, the total number of ways to arrange the books is:
4! * 2! * 3! = 24 * 2 * 6 = 288 ways.
To solve this problem, we need to consider the groups of books as distinct units. Let's treat the algebra books as one unit, the geometry books as another unit, and the pre-calculus books as a third unit. We need to find the number of ways these three units can be arranged on the shelf.
First, consider the three book units as one unit. So now, we have a total of three units to arrange on the shelf: the combined unit of algebra books, the combined unit of geometry books, and the combined unit of pre-calculus books.
The number of ways to arrange these three units is simply the number of permutations of three objects. The number of permutations of 'n' objects is given by n!.
In this case, we have three units to arrange, so the number of ways to arrange the three units is 3!.
3! = 3 x 2 x 1 = 6
Therefore, the teacher can arrange the books on the shelf in 6 different ways if the books covering the same subject matter are kept together.