Distribute 14 books to 2 people so that each has at least 3 books.

Question

I've been trying to solve a problem, but the solution I got looks extrmely unlikely to be right. Perhaps some can point where I'm wrong. The problem is the following.

We have $14$ different books. We want to distribute them to person $A$ and $B$ in a way that none of them are left with less than $3$ books. I've tried the following.

Let $L=\{1, 2, 3, ..., 14\}$ be the set of books in my possesion.

$A$ and $B$ must receive at least $3$, and the $3$ books that I'll give to any of them will be a subset of $L$. There are $\binom{14}{3}$ subsets of $3$ elements of $L$, so when I give the first $3$ books to the first person (whether its $A$ or $B$) I have $\binom{14}{3}$ different ways to do it. By the time I'll give the second person other $3$ books I will have only $11$ books remaining, cause I gave $3$ to the first person. The $3$ books I'll give to the second person will be subsets of $L'$, the set of the $11$ books remaining, so I'll have $\binom{11}{3}$ different ways of giving the second person its three books.

We have to cases:

$a)$ First person is $A$, in which case I have $\binom{14}{3}\binom{11}{3}$ possibilities, the first three books for $A$ and three books of the remaining ones for $B$.

$b)$ First person is $B$, in which case I have $\binom{14}{3}\binom{11}{3}$ possibilities, the first three books for $B$ and three more books for $A$.

Together, this adds up to $\binom{14}{3}\binom{11}{3} + \binom{14}{3}\binom{11}{3}$ possibilities.

But now that I made sure that each person gets at least $3$ books, I have $8$ remaining books, that I can give to anyone I want however I want. So my first decisition is give $1$ of $8$ books to $A$ or $B$; second is to give $1$ of $7$ books to $A$ or $B$, and so on. So I have $8!+8!$ possibilities here. At last, I have

$[\binom{14}{3}\binom{11}{3} + \binom{14}{3}\binom{11}{3}]*(8!+8!) = 2[\binom{14}{3}\binom{11}{3}]*2(8!)=9686476800$.

I'm no genius but I don't think that giant number could be the right answer.

Answer 1

Let A and B be the people among whom the 14 books (all different) will be distributed, with the condition that each of these people must receive at least 3 books.

Obviously, the order in which the books are distributed between A and B is indifferent.

A very easy way to visualize the process of distribution of these 14 books is to imagine that each book is affixed with a label with the letter A or B. Each book will be assigned to the person whose name matches the one of the labels and the number of different ways of performing this labeling is

$${{n}_{1}}\equiv {{2}^{14}}=16384$$

To fulfill the condition that A and B receive at least 3 books each, we must eliminate those labeled with less than 3 "A" or less than 3 "B". The number of labeled with 0, 1 or 2 "A" is the same as the number of labeled with 0, 1 or 2 "B".

• Number of labels A in a labeled = 0, (14 B) $\Rightarrow $ Number of labeled:

$${{n}_{2}}\equiv 1.$$

• Number of labels A in a labeled = 1, (13 B) $\Rightarrow $ Number of labeled:

$${{n}_{3}}\equiv 14.$$

• Number of labels A in a labeled = 2, (12 B) $\Rightarrow $ Number of labeled:

$${{n}_{4}}\equiv \left( \begin{matrix} 14 \\ 2 \\ \end{matrix} \right)=91.$$

Therefore, the number of ways to make the proposed distribution is,

$$n={{n}_{1}}-2\left( {{n}_{2}}+{{n}_{3}}+{{n}_{4}} \right)=16384-2\left( 1+14+91 \right)$$

that is,

$$n=16172.$$

Answer 2

Here the question is ambigious Are all the books identical or distinct? I assumed them to be identical Give 3 books to each A andB Now we are left with 8books to distribute let a books be given to A and b books be given to B Therefore; a+b=8
The no. Of non-negative integral solutins to above equation will be the required no. Of ways So; it is 9

Answer 3

There are $2^{14} = 16\,384$ ways to distribute $14$ distinct books to $2$ distinct people, disregarding the minimal count requirement.

Of these, there are

• $\binom{14}{0} = 1$ ways to give A no books,
• $\binom{14}{1} = 14$ ways to give A exactly one book, and
• $\binom{14}{2} = 91$ ways to give A exactly two books,

and identical ways to give B exactly $0$, $1$, or $2$ books, for a final count of $(1+14+91)\times2 = 212$ ways for books to be distributed in rule-breaking ways. This leaves $16\,384-212 = 16\,172$ ways to distribute the books.