We have $3$ books and $3$ shelves. We are to put $2$ books on $1$ shelf and $1$ book on the other two.

Answer given in the book is $6$ but I feel that—— We can select $2$ books from $3$ in $C_{(3,2)}=3$ ways and then we can arrange these two selected books in $2!=2$ ways. Then the remaining book can be put on the either of the remaining shelves and hence this can be done in $2$ ways. So in my view total number of ways should be $3\times 2\times 2=12$ not $6$. Where am I making the mistake or is it that answer given in the book is wrong? Thanks in advance to anyone who will help. 🙏

  • $\begingroup$ @N.F.Taussig Just seen this comment of yours. I believe you are right that if books are identical then answer should be 6. This is my thought process— 2 books are to be put on one of the shelf. We can pick one shelf in 3 ways. Now remaining book can be put on one of the remaining shelves in 2 ways. Hence answer=3*2=6. Is this right process or wrong? $\endgroup$ – Akash Apr 22 at 11:42
  • $\begingroup$ You have correctly solved the case in which the books are identical. However, I now realize that you solved the case in which the books are different incorrectly. Your oversight was that you did not choose which shelf would receive two books. I apologize for initially telling you that you had solved that problem correctly when you had not. $\endgroup$ – N. F. Taussig Apr 22 at 20:33

The answer depends on whether the books are identical or different.

In how many ways can three identical books be placed on three shelves if two books are placed on one shelf and one book is placed on another shelf?

As you observed in the comments, the shelf that receives two books can be selected in three ways and the shelf that receives the remaining book can be selected in two ways. Hence, the books may be distributed in $3 \cdot 2 = 6$ ways.

If we further observe that once we have chosen which shelf will receive two books and which shelf will receive one book, there is only one way to choose which shelf will receive no books, we obtain the answer posted by Eureka.

In how many ways can three different books be placed on three shelves if two books are placed on one shelf and one book is placed on another shelf?

We must choose which two of the three books are placed on the same shelf, choose which of the three shelves receives those books, arrange the two books on that shelf, and choose which of the other two shelves receives the remaining book. Hence, there are $$\binom{3}{2}\binom{3}{1}2!\binom{2}{1} = 36$$ such distributions.

In your attempt, you forgot to choose which shelf would receive two books.

  • $\begingroup$ I’m satisfied now with this. $\endgroup$ – Akash Apr 23 at 7:56
  • $\begingroup$ I’m self studying this topic. There is another doubt I want to clear up: Say we have x identical objects. And we have to select y (<x) of them. It can be done in 1 way only, right? Because they are identical. $\endgroup$ – Akash Apr 23 at 7:59
  • $\begingroup$ My last question to you(I’m really grateful for your help 🙏).- Find no. Of ways in which 4 identical balls can be placed in 6 identical boxes, if not more than one ball goes into a box? I feel it can be done in one way only because balls and boxes are identical and 4 of boxes receive 1 ball and rest two receive none. But answer bizarrely is given as C(6,4)=15. $\endgroup$ – Akash Apr 23 at 8:03
  • $\begingroup$ You are correct that there is only one way to select $y$ of $x$ identical objects for $y \leq x$. The number $\binom{6}{4}$ represents the number of ways of selecting four of the six boxes in which to place a ball. As you observed, for this answer to make sense, the boxes must be distinguishable, if only by their positions. If the boxes are indistinguishable, then there is only one such distribution, as you observed. $\endgroup$ – N. F. Taussig Apr 23 at 9:43
  • $\begingroup$ Thank you very much, Sir. I felt that I had learnt this concept to some extent but these wrong answers were causing me headache. Thankfully something good came out of it as I learned something new. Thanks again and good day. 🙏 $\endgroup$ – Akash Apr 23 at 11:49

I think that the order of the $2$ books on the same shelf doesn't matter so you have to divide for $2!$. The best procedure would be this. Notice that we have there will be always a shelf with $2$ books, $1$ with $1$ book and $1$ with $0$ books so we must permutate:


That is banally:



  • $\begingroup$ But books are different and so the order of 2 books on 1 shelf will/should matter. In my view, which shelf is empty and which one is filled, doesn’t that make up a distinct permutation? Say 210 and 201? $\endgroup$ – Akash Apr 22 at 8:28
  • $\begingroup$ @N.F.Taussig Thanks. 👍 $\endgroup$ – Akash Apr 22 at 11:39

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