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# In How Many Different Ways Can 5 Books Be Arranged On A Shelf?

## In how many different ways can you arrange 7 books on a shelf.

Because you cannot put the same book in two places at once, the answer is 7! (the ! is read as factorial).7! = 7*6*5*4*3*2*1 = 5040

## In how many different ways can you arrange 7 books on a shelf.

There are 7 possibilities for the first position,then 6 remaining books for the next spot,5 for the one after, and so on.7x6x5x4x3x2x1 permutations.

## how many ways can 5 red, 3 green, and 4 blue books be arranged on a shelf if….

..only the red and blue books must be together?Please show/explain your calculations! ðŸ™‚

The red and blue books need to be together so start by figuring out where these will be. There are a total of 5+4 = 9 red and blue books.Think of the places to put these books as being 9 slots on the shelf._ _ _ _ _ _ _ _ _You need to choose 5 places out of 9 to put the red books. The remaining 3 slots will be where the blue books go.There are 9c5 = 126 ways to order the red and blue colors.Now consider the green books. You need to place the red and blue books (in one group) somewhere amid the green books (or on an end)._G_G_G_There are 4 choices of where to put the red and blue books among the green books.There are a total of 4*126 = 504 ways to order the colors so that the red and blue books are together.At this point you have to decide on something not really given in the question. If all the books that are the same color are exactly the same book then you are done. The answer is 504. If the books that have the same color have different titles then you have to do some more work.You already figured out the combinations possible for color order. Now you just need to figure out the orders for the titles.You can place the 5 red books in 5! different orders.Similarly there are 3! for green and 4! for blueThe total number of ways you can arrange the books on the shelf is then504*5!*4!*3! = 8,709,120

## In how many different ways can 6 books be arranged on a shelf if one of the books, book X, has to be first.

Suppose all 6 books are different (i.e. no repetition) and that there are no other constraints.First book -> XSecond to sixth book can be any of the 5 books.This means I have 5 choices for the second book, 4 choices for the third book, 3 choices for the fourth book, 2 choices for the fifth book, and 1 choice for the sixth book.Thus, number of ways= 5 * 4 * 3 * 2 * 1= 120 ways

## How many ways can six different books be arranged on a shelf.

6 books can be arrange in ;6! ways i.e.6*5*4*3*2*1 = 720 ways

## In how many different ways can someone arrange 6 books on a shelf.

In how many different ways can someone arrange 6 books on a shelf?

Millions of ways. Some right side up, some sideways, some open, some closed, etc. ….But if you mean all right side up in the normal way, then there are 6 choices for the first book, 5 for the second, etc. 6 x 5 x 4 x 3 x 2 x 1 = 720 ways.

## how many ways can six different books be arranged on a shelf.

I think it is 7206!6*5*4*3*2*1=720not sure if this is what you meant.Factorials are used in combinatorics. For example, there are n! different ways of arranging n distinct objects in a sequence. (The arrangements are called permutations.) And the number of ways one can choose k objects from among a given set of n objects (the number of combinations), is given by the so-called binomial coefficient

## In how many different ways can 6 books be arranged on a shelf.

In how many different ways can 6 books be arranged on a shelf?(a) 6(b) 120(c) 720

The first book can be any one of the 6 books.The next book can be any one of the remaining 5.And so on.6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

## In how many different ways can 5 books be arranged on a shelf.

A. 5B. 25C. 120Thanks for helping! ðŸ™‚

5!=1*2*3*4*5=120 C

## 45 thoughts on “In How Many Different Ways Can 5 Books Be Arranged On A Shelf?”

1. 3 2 1 6 5 41

2. 5!=1206! = 6*5*4*3*2*1 = 30*4*3*2*1 = 120*3*2*1 = 360*2*1 = 720*1 = 720 ways

3. 4) alphabetically Z-A5! = 120

4. 2 1 6 5 4 3the way to arrange 4 blue books = P(4,1) = 4you can arrange the books on this condition = 5! * 4! * 51 = 120 * 24*120 = 273600 ways

5. red and blue half to be touching

6. 2 3 4 5 6 1124 5 6 1 2 3

7. you do 5x4x3x2x1 because first you have 5 options of which book goes first. when you put on book on the shelf you only have 4 choices left, etc

8. 300?

9. 720.

10. Is this a Permutation or an Combinationa million.) if the chocolates are pulled out at as quickly as than the danger is 5/20 * 4/19 = 20/380 = a million/19, 5/20 * 6/19 = 30/380 = 3/38 2.) 7/sixteen * 6/sixteen * 5/sixteen = 210/4096 = one hundred and five/2048 So On So forth

11. 5! = 5 * 4 * 3 * 2 * 1 = 120

12. You can arrange 6 books on a shelf 6*5*4*3*2*1=720times6!= 6*5*4*3*2*1= 720

13. (0 pts) 720 ways5! ways

14. (2 pts) 5,040 ways

15. Does throwing them in the air count as arranging them?

16. 1 2 3 4 5 6the way to arrange 5 red books together, 4 blue books together and 3 green books : P(5,1) = 5!(0 pts) 28 ways6*5*4*3*2*1=720 ways.

17. [DELDUP()]

18. 9. In how many different ways can you arrange 7 books on a shelf? (2 points)X, then…5 ! ways5 6 1 2 3 4Example you place two books vertically on the sides and the rest horizontally on top of each other in between. Arrange all vertically or horizontally etc there are just so many ways one can arrange 6 books.

19. c

20. ( note :the number of group = 1 red + 1 blue + 3 green = 5)

21. 3) alphabetically A-Z

22. (just my guess)

23. The ways that you may organize 6 books on a shelve is 6! = 720 Reasoning in the back of that is for the first empty house : there are 6 feasible books that you may put in there 2nd empty area, u can best decide on out of 5 books to place there (considering that u’ve already placed one booklet in the first slot) proceed utilizing this logic with the rest books and you are going to get 6 * 5 * 4 * three * 2 * 1 = 6! = 720

24. well it would still be 60 ways as the guy above said

26. 6) by author720 not including alternative arrangements by one upsidedown, one sideways, etc..

27. 3 4 5 6 1 25 x 3 x 4 = 60 ways

28. Alrighty Chief. Like the guy below. color, alphebetically, title, author, publishing date, date you got them, size and probably more.

29. 6! + however many if you wanted to rotate/stack them. Possibly infinite.if you have to put all book on shelf here how4 3 2 1 6 56 5 4 3 2 1I got 30 cause 6×6 is 36 minus the original arrangement.

30. if you need more answers let me know…I just took this test :)the way to arrange 5 red books = P(5,1) = 5!(0 pts) 823,543 ways5 4 3 2 1 6[/DELDUP]6 1 2 3 4 5

31. 1. 120=120 ways

32. 1) by size-larger books left to right

33. 2 /2 points

34. This is a permutation because the order does matter.

35. 7) by publisher5) by subject120 ways !

36. 6-1=5

37. 6*5*4*3*2*1 = ? find out.

38. alphabetically,by color by pages bysize or just say 6 different ways

39. 6! 1*2*3*4*5*6 = 720

40. These people confuse me with their big numbers.Put the book that should be the first in position 1. The other 5 books can be permuted in 5! = 120 ways.

41. 2) by size-smaller books left to right1 6 5 4 3 2