Sagot :
The number of ways of the books can be arranged are illustrations of permutations.
- When the books are arranged in any order, the number of arrangements is 3628800
- When the mathematics book must not be together, the number of arrangements is 2903040
- When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
- When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400
The given parameters are:
[tex]\mathbf{Novels = 3}[/tex]
[tex]\mathbf{Mathematics = 2}[/tex]
[tex]\mathbf{Chemistry = 5}[/tex]
(a) The books in any order
First, we calculate the total number of books
[tex]\mathbf{n = Novels + Mathematics + Chemistry}[/tex]
[tex]\mathbf{n = 3 + 2 + 5}[/tex]
[tex]\mathbf{n = 10}[/tex]
The number of arrangement is n!:
So, we have:
[tex]\mathbf{n! = 10!}[/tex]
[tex]\mathbf{n! = 3628800}[/tex]
(b) The mathematics book, not together
There are 2 mathematics books.
If the mathematics books, must be together
The number of arrangements is:
[tex]\mathbf{Maths\ together = 2 \times 9!}[/tex]
Using the complement rule, we have:
[tex]\mathbf{Maths\ not\ together = Total - Maths\ together}[/tex]
This gives
[tex]\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}[/tex]
[tex]\mathbf{Maths\ not\ together = 2903040}[/tex]
(c) The novels must be together and the chemistry books, together
We have:
[tex]\mathbf{Novels = 3}[/tex]
[tex]\mathbf{Chemistry = 5}[/tex]
First, arrange the novels in:
[tex]\mathbf{Novels = 3!\ ways}[/tex]
Next, arrange the chemistry books in:
[tex]\mathbf{Chemistry = 5!\ ways}[/tex]
Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.
Literally, the number of books now is:
[tex]\mathbf{n =Mathematics + 1 + 1}[/tex]
[tex]\mathbf{n =2 + 1 + 1}[/tex]
[tex]\mathbf{n =4}[/tex]
So, the number of arrangements is:
[tex]\mathbf{Arrangements = n! \times 3! \times 5!}[/tex]
[tex]\mathbf{Arrangements = 4! \times 3! \times 5!}[/tex]
[tex]\mathbf{Arrangements = 17280}[/tex]
(d) The mathematics must be together and the chemistry books, not together
We have:
[tex]\mathbf{Mathematics = 2}[/tex]
[tex]\mathbf{Novels = 3}[/tex]
[tex]\mathbf{Chemistry = 5}[/tex]
First, arrange the mathematics in:
[tex]\mathbf{Mathematics = 2!}[/tex]
Literally, the number of chemistry and mathematics now is:
[tex]\mathbf{n =Chemistry + 1}[/tex]
[tex]\mathbf{n =5 + 1}[/tex]
[tex]\mathbf{n =6}[/tex]
So, the number of arrangements of these books is:
[tex]\mathbf{Arrangements = n! \times 2!}[/tex]
[tex]\mathbf{Arrangements = 6! \times 2!}[/tex]
Now, there are 7 spaces between the chemistry and mathematics books.
For the 3 novels not to be together, the number of arrangement is:
[tex]\mathbf{Arrangements = ^7P_3}[/tex]
So, the total arrangement is:
[tex]\mathbf{Total = 6! \times 2!\times ^7P_3}[/tex]
[tex]\mathbf{Total = 6! \times 2!\times 210}[/tex]
[tex]\mathbf{Total = 302400}[/tex]
Read more about permutations at:
https://brainly.com/question/1216161