The number of ways in which 5 different books can be given to 10 people with a condition

Question

The number of ways in which 5 different books can be distributed among 10 people if each person can get at most two books is?

My Attempt:

The number of possibilities is
$(2,2,1)$[rest all zero],$(1,1,1,1,1)$[rest all zero], $(2,1,1,1)$[rest all zero]

Therefore, dividing 5 books and distributing them will give me,

$\binom{10}{3}[\frac{5!}{2!2!1!2!}] + \binom{10}{5}[\frac{5!}{1!1!1!1!1!}] + \binom{10}{4}[\frac{5!}{2!1!1!1!}]$

This may be the answer according to me.

But I don't know the correct answer to the given question. So, can anyone check my work?

Answer 1

Your approach is right. But suppose we are looking at the case $(2,2,1)$, then once you have chosen three people over the ten, say people $p_1,p_2,p_3$ you have to choose which one of the three receives 1 book (i.e. multiply by three), then you have to choose which books are to be given to person 1 (choose two over five) and to person 2 (choose two over three). This will cover all cases in this setting.

$${10\choose 3}\times 3\times {5\choose 2}\times{3 \choose 2}.$$

The other two cases $(2,1,1,1)$ and $(1,1,1,1,1)$ are similar, I'll leave you to it.

Answer 2

test case for {2,1,1,1}

no of ways to select 4 person($P_1,P_2,P_3,P_4$) =$\binom{10}{4}$

no of ways we can divide 5 books into 4 groups of {2,1,1,1}($G_1,G_2,G_3,G_4$)

=$\frac{5!}{2!1!1!1!}$

now we have to assign these ($G_1,G_2,G_3,G_4$) to ($P_1,P_2,P_3,P_4$) which can be done in 4 ways

Total = 4 $\binom{10}{4}$$\frac{5!}{2!1!1!1!}$ = 240 $\binom{10}{4}$

Method 2:

instead of dividing into groups lets calculate just how many ways we can choose 2 books out of 5 =$\binom{5}{2}$

Now these can be arranged to ($P_1,P_2,P_3,P_4$) in 4! ways

Total = 4! $\binom{5}{2}$ $\binom{10}{4}$ = 240 $\binom{10}{4}$

You can choose any method you want , its just that in method one you didnt consider the matter of assigning

Also when @Tal-Botvinnik mention (2 over 5) and (2 over 3) he is choosing {2,2,1} by method 2

Answer 3

You have correctly identified the cases.

Five people each receive one book: There are $\binom{10}{5}$ ways to select which five people will receive a book and $5!$ ways to distribute the five books to those people. Hence, there are $$\binom{10}{5}5!$$ such distributions.

You have calculated this case correctly.

One person receives two books and three others each receive one book: There are ten ways to choose which person receives two books and $\binom{5}{2}$ ways to select the books this person receives. There are $\binom{9}{3}$ ways to choose three of the other nine people to receive one book each and $3!$ ways to distribute the remaining three books among those people. Hence, there are $$\binom{10}{1}\binom{5}{2}\binom{9}{3}3!$$ such distributions.

In your attempt, you selected which four people would receive a book and calculated the number of ways of arranging the five books into a group of two and three single books. However, you forgot to choose which of the four recipients would receive two books. Therefore, you need to multiply your result by $4$.

Two people each receive two books and a third person receives one book: There are $\binom{10}{2}$ ways to choose which two people receive two books. There are $\binom{5}{2}$ ways to select which two books will be given to the younger of those two people and $\binom{3}{2}$ ways to select which two of the remaining three books will be given to the older of those people. There are eight ways to select which of the remaining eight people receives the remaining book. Hence, there are $$\binom{10}{2}\binom{5}{2}\binom{3}{2}\binom{8}{1}$$ such selections.

In your attempt, you selected which three people would receive a book, but you did not choose which two of them would receive two books, so you need to multiply by $\binom{3}{2}$. Also, the number of ways of arranging the five books into two groups of $2$ and one group of $1$ should be $\frac{5!}{2!2!1!}$. You have an extra factor of $2!$ in your denominator.