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Sagot :
Answer:
There are 40,320 ways, in which 8 books can be arranged on a shelf.
Solution:
Here, we are to find the number of ways in which 8 books can be arranged on a shelf. The total number of books is 8 and the way of arranging books is also 8.
- If one book is placed in the first place, then 7 books will be placed in front of it. If 2 books are placed in the 2nd place, then only 6 books can be placed after that book. This sequence will continue till 1 .
Permutations :
- A permutation is an arrangement of objects in a definite order.
➲ P ( n, r )= n ! / ( n - r ) !
- n = total number of objects
- r = number of objects selected
The number of ways to arrange 8 books on a shelf will be :
➝ P ( n, r ) = n ! / ( n - r ) !
➝ P ( n, r ) = 8 ! / ( 8 - 8 ) !
➝ P ( n, r ) = 8 ! / 0 !
➝ P ( n, r ) = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 / 1
➝ P ( n, r ) = 40, 320
ㅤㅤㅤㅤㅤㅤ~ Hence, there are 40,320 ways in which 8 books can be arranged on a shelf !
If the 3 books can't be together (but any 2 of the 3 can be) then it's 36,000 ways. If any one of the 3 books can't touch either of the other 2, it's 15,840 ways.
Explanation:
I'm not sure if the question is saying that all three books can't be together in a group (and so biology and history can be together, but biology, history and programming can't), or if the books in that group can't sit next to other books in the group (and so biology and history can't be together). Let's do it both ways - case 1 will be all three together and case 2 will be any 2 together.
Before we work this, let's first see the number of ways we can have all 8 books arranged. Order matters and so we can approach this using a permutation equation. Moreover, since we're using all the books, we'll end up with the total number being:
8 ! = 40 , 320 .
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