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