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Q.The number of ways, we can arrange 5 books in 3 shelves ________.
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why are we taking permutation, and not combination

generally formula is $_{r-1}^{n+r-1}\textrm{C}$

right?
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We take combination in case of unordered arrangements. But here, the arrangements are ordered, since books are distinct, not the same. Had the book been same, we would take combination, since then the arrangements would be unordered.
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3 Answers

11 votes
11 votes
Best answer

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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4 Comments

 sir can u add one question with solution to simply define all these cases ... my probability of correcting these questions is appox 0.5 sometimes i  am in the wrong direction but dont even know that i am in wrong direction

plzz add comment after this so we will get notification : )

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i was having doubt so i verified it with the technique which @Pragy Agarwal sir gave and solution given by @balchandar reddy san gave 
for 5 books and instead of 3 shelf i took 4 and in both the cases 6720 is the answer
Thank u for simplifying this question.

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@Pragy Agarwal sir u said 3^5 will count no. of ways of putting 5 books in 3 shelves then what will                             

   n+r-1 C r-1    will count here as both are calculating different answers and by this formula putting                n=5 , r =3  here we get i.e. 5+3-1C3-1 = 7C2 = 21 and then if we multiply this selection cases by 5! as arrangement also need to be considered in this book arrangements problem then we are getting correct answer = 2520.

 

so I think using 3^5 is not correct here .why? 

 

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3 votes
3 votes

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

edited by

4 Comments

@balchandar reddy san

suppose consider the arrangement that u mentioned in your comment above, i.e. S1 contains A,B,C , S2  contains D and S3 contains E. 

now will this arrangement i.e. A, B,C contained in shelf S2, D contained in S1 and E contained in shelf S3 be a different arrangement than the previous one?? Shouldn't this arrangement be distinct?  We are not considering the shelves as identical right? 

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Ok..let me explain it by taking one more example..with 2 books and 2 shelves where i want to put 1 book in each shelf

S1,S2

1,1(A,B)

You know that this can be done only in 2 ways, but according to my way of doing, it would be

2C1*2*1C1*1 = 4 Cases, because 

Case 1: I chose A first and place it in S1 and chose B next and place it in S2 {A->S1, B->S2}

Case 2: I chose A first and place it in S2 and chose B next and place it in S1 {B->S1, A->S2}

Case 3: I chose B first and place it in S1 and chose A next and place it in S1 {B->S1,A->S2}

Case 4: I chose B first and place it in S2 and chose A next and place it in S1 {A->S1, B->S2}

As you can see, even though there are four cases, there will be only 2 distinct arrangements..this happens only when the two shelves contains same number of elements.. so it should be 2C1*2*1C1*1 / 2!

and similarly for 3 books and 3 shelves if i have to put 1 book in each

S1,S2,S2

1,1,1(A,B,C)

3C1*3*2C1*2*1C1*1 / 3! = 36 / 3!  as all three shelves have same number elements we are dividing here with 3!..

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finally got it..thanks :)
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1 vote
1 vote
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.

1 comment

Can you please explain what does bold factorial 3 and 2 represent? I am not able to get it. (5C3*3*2) *3!*2!

Please help and clarify.

​​​

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