How many ways can 20 books be placed on 5 shelves?

= 23×22×21 = 10626. This is a question of combinations. we get: Number of combinations= 20!/15!. 5!

How many ways can 10 books be arranged on a shelf if one of the books must be on one end?

question_answer Answers(1) ways. ∴ hence the total number of arrangements = 9!

How many ways are there to order 5 books on a shelf?

120 ways
That’s 5! So, there are 120 ways to arrange five books on a bookshelf.

How many ways are there to place 7 books in order if a certain book must always come first and another certain book must always come last?

F. A. All the seven books are distinct or they are different. So it’s just seven victoria ways of arranging them, which means this will be 5040.

How many ways can we place 7 books on a bookshelf?

= 5040 ways. Hence we can arrange 7 different books on a shelf in 5040 ways.

How many different books can be distributed on 5 shelves?

You have to distribute 3 books on 5 shelves so possibilities are (0,0,1,1,1), (0,0,0,1,2), (0,0,0,0,3)= and shelves are also different use the group distribution formula then *5! You get answer 210 A shop assistant is going to put 4 different English books, 5 different Chinese books, and 3 different math books in a bookcase which has 3 shelves.

How do you divide the books among the shelves?

First of all, there are 3 ways to divide the books among the shelves: 1) all 3 on one shelf; 2) 2 on one shelf and 1 on another; 3) each book on a separate shelf. And for any given arrangement among the shelves, there are 3! = 6 ways of distributing the books in the given “slots”.

Where do you place Book 5 on the shelf?

Place book 5 on the shelf, it can go far left, between the left two books, in the middle, between the right two books or on the far right (e.g. _ 1 _ 4 _ 3 _ 2 _ ), there are 5 ways to do this for each arrangement of 1,2,3 and 4.

How many ways in which 5 books can be placed?

Fourth book can be placed in either of the 2 remaining places, leaving 1 for the final book. So total number of ways in which 5 books can be placed is: 5*4*3*2*1 =5! = 120. With all books upright: 5x4x3x2x1=120 (spine facing out). With all books upright: 5x4x3x2x1=120 (spine facing in).