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Q.The number of ways, we can arrange 5 books in 3 shelves ________.
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As @aambazinga mentions, the answer is 7! / 2! = 2520

There are 5 books (nothing has been mentioned about whether the books are different or identical, so we always assume different). There are 3 shelves.

You can think of arranging the books into 3 shelves as having 5 markers for books, and 2 markers for the dividers of the shelves.

example: 12 | 345 | 67

So, you have 7 objects in total, 2 of which are identical (the dividers)

The number of ways of arranging these is $7! / 2!$


Why is $3^5$ incorrect? Understanding this is as important as being able to correctly answer the question.

$3^5$ says that each of the 5 books can go in any of the 3 shelves. This is correct, but it counts the ways of "putting" the books in the shelves, and not "arranging" them. For "arranging" them, we also need to count the number of ways the books can be arranged (permutations), once they're placed in the shelves.

That would be a longer calculation, as @balchandar reddy san did, and will finally lead you to the correct answer.

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S1,S2,S3            W/O Arrangement                         With arrangement  

5,0,0                      3  ( any of the three shelfs)                (3) *5!

4,1,0                      $\binom{5}{4}$*3*$\binom{2}{1}$                                       (5C4*3*2) *4!

3,2,0                      $\binom{5}{3}$*3*$\binom{2}{1}$                                       (5C3*3*2) *3!*2!

3,1,1                      $\frac{\binom{5}{3}*3*\binom{2}{1}*2 * \binom{1}{1}*1}{2} $                      ( 5C3*3*2C1*2 / 2) *3!

2,2,1                      $\frac{\binom{5}{2}*3*\binom{3}{2}*2 * \binom{1}{1}*1}{2} $                        (5C2*3*3C2*2 / 2) *2!*2!

Total:                    243                                                 2520

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Assuming the books are distinguishable , and so are the shelves.

There can be 5! arrangements from left to right of the books,

Example : 12345 or 34512 or 34125 ,etc.

Now assuming you have made an arrangement of books say 12345 , the problem is now partitioning the books into 3 shelves, since we know, to partition into k partitions we need k-1 bars. Example : 12|3|45 or 123| |45. Therefore now this is the same problem of dividing n objects into k partitions. Now there are 5+2=7 objects and we need to select the position of the bars. We can do so in $\binom{7}{2} ways$. Therefore total arrangements=5!*7!/(5!*2!)=7!/2!=3*4*5*6*7=2520 ways.

 

Please comment if something is not clear.

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