5 books can be lined up on a shelf in (5 x 4 x 3 x 2 x 1) = 120 different sequences.
30 ways.
There are only 5 places on the shelf. You have 7 books to choose from. We will ignore the order of the books on the shelf. The first place can be filled from a choice of 7 books, the next place from 6, the next place from 5, the next from 4, and the last of the 5 places from 3 books. So the number of ways of choosing the 5 is found from 7 * 6 * 5 * 4 * 3 = 2520
Consider one placement at a time. The first book on the shelf could be any of seven. The second could be any of the remaining six. This continues with the rest. The total combinations are: 7! = 7*6*5*4*3*2*1 = 5040
The answer depends on how many books on each subject there are.
792 different groups of 5 books, in 95,040 different sequences.
asdasdadadadasdasd
because of gravity. i think
30 ways.
Alternate them.
Alternate them.
The answer to this one is 24. You can do this mathematically by 4*3*2*1.
There are only 5 places on the shelf. You have 7 books to choose from. We will ignore the order of the books on the shelf. The first place can be filled from a choice of 7 books, the next place from 6, the next place from 5, the next from 4, and the last of the 5 places from 3 books. So the number of ways of choosing the 5 is found from 7 * 6 * 5 * 4 * 3 = 2520
There are 5! (that is, 120) distinct ways to arrange five items. Only 1 of them will have the books in alphabetical order by title. So the probability that it happens by random is 1/120.
Yes. They are real.
120. You do 5*4*3*2*1=120. you multiply the number that you are given for example how many times can books 3 be arranged on a shelf you multiply 3*2*1=6 that is your answer
The answer would be 7! or (7*6*5*4*3*2*1)=5040
Choose 3 then 2 then 1; 3*2*1 = 6 ways.