Eight different books are to be packed in eight different bags. If two of the bags are too small to hold five of the books, in how many ways can the books be packed?
Answer - 4320
My Problem : I can not understand the line "two of the bags are too small to hold five of the books."
Can anyone make me understand this line?
On
The phrase "two of the bags are too small to hold five of the books" means probably that you cannot put any of those $5$ books in any of those $2$ bags.
The question also mentions that $8$ books need to be put in $8$ bags, that in turn requires all bags to have exactly one book.
For those two particular bags, we can only put $2$ out of $3$ smaller books available. This is done in $\binom{3}{2}2!$ ways. For the remaining $6$ books we may put them in $6!$ ways in the remaining bags.
So the number of solutions is $\binom{3}{2}\cdot 2!\cdot 6! = 4320$.
Assuming 1 book per bag.
You have $8$ different bags $\{1,2,3,4,5,6,7,8\}$ and $8$ different books $\{A,B,C,D,E,F,G,H\}$.
Each bag can hold at most one book and bags $1$ and $2$ are too small to hold books $A,B,C,D$ and $E$.
This means books $A$ to $E$ can only go in bags $3,4,5,6,7$ and $8$. There are $6!/1!$ ways to place these, each way leaving one empty bag out of those six. Then for each way there are $3!$ ways to place books $F,G$ and $H$ in the remaining three bags (bags $1, 2$ and the empty bag left over from bags $3$ to $8$).
$$\frac{6!}{1!}\cdot 3!=4320\tag{Answer}$$
To illustrate the restrictions of the question check out this table:
$$\begin{array}{cc} &\text{Bags}\\ \text{Books}& \begin{array}{c|c|c|c|c|c|c|c|c|} &1&2&3&4&5&6&7&8\\\hline A&\textbf{X}&\textbf{X}&&&&&&\\\hline B&\textbf{X}&\textbf{X}&&&&&&\\\hline C&\textbf{X}&\textbf{X}&&&&&&\\\hline D&\textbf{X}&\textbf{X}&&&&&&\\\hline E&\textbf{X}&\textbf{X}&&&&&&\\\hline F&&&&&&&&\\\hline G&&&&&&&&\\\hline H&&&&&&&&\\\hline\end{array}\end{array}$$
Of course, just to reiterate, we must also remember that this is with the additional assumption of 1 book per bag.