Question

In how many ways can $3$ books be given to $5$ person if,
(i) no person can have more than one book?
(ii) a person can have any number of books?

Answer 1

Consider all are different books,

Given that there are $3$ books and $5$ people to be distributed,

$1)$ no person can have more than one book,

Now, consider the people as gaps,

so $3$ different books should be filled in $5$ gaps such that each can contain only one book,

$\therefore$ number of ways of doing this is ${}^5{P}_3=60$,

i.e., the empty gaps do not receive any book.

$2)$ a person can have any number of books

Case 1 : no person can have more than one book

the same case mentioned above which are $60$ ways.

Case 2 : one person receiving $2$ books,

As all the books are different the number of ways of selecting $2$ books out of them are ${}^3{C}_2=3$

So those can be distributed in $5$ ways i.e., to any of the $5$ people and the other book can be distributed to any of the remaining $4$ people.

$\therefore$ total number of ways are $\mathrm{3.5.4}=60$

Case 3 : all the three give to a single person,

we can do that in $5$ ways because we can give all of them to any of them

$\therefore$ total number of ways of distributing $3$ different books to $5$ people such that each one can receive any number of books is $60+60+5=125$