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Assuming they are all right side up and spine out, there would be 40,320 different ways of arranging them.
Consider: There are 8 choices for the 1st book, 7 for the 2nd, 6 for the 3rd, 5 for the 4th, 4 for the 5th, 3 for the 6th, 2 for the 7th and 1 for the 8th. Multiplying these numbers out:
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320. More than you thought?
The mathematical name for this is called finding the factorial of a number!
It is written n! ( n followed by an explanation point) -
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How many ways can 7 books be arranged vertically on a bookshelf?
If the books have to be the correct way up and spine outwards: 7! ways =7x6x5x4x3x2x1 =5040 ways. If the books can be any way in (upside down, spine inward, etc.): (7!x4^7) ways =7x4x6x4x5x4x4x4x3x4x2x4x1x4 =82,575,360 ways
How many ways can seven books be arranged on a shelf if one of the books is an algebra text and it must be in the center?
it depends on the shelf of course
How many ways can the letters of the word meddles be arranged?
In how many distinct ways can the letters of the word MEDDLES be arranged?
How many ways can 7411111 be arranged?
The answer is 7!/5! = 42 ways.
How many different ways can ten books be arranged on a shelf?
Using the binomial coefficient and the 'choose' function, 10 books can be arranged 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800 different ways.
