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

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.

• Are all the books to be distributed? If so, then the collection of books that goes to $A$ determines those that go to $B$. Thus the answer is just $\sum_{i=3}^{11}\binom {14}i=16172$ – lulu Apr 15 at 17:09
• That number is indeed too big. You should get something less than $2^{14} = 16384$. You have counted many cases multiple times – Henry Apr 15 at 17:09
• The reason this overcounts is that the same distribution can be reached by multiple different decision paths in the above scheme. Suppose we ultimately give (a specific set of) 8 books to $A$, and the other books to $B$. Any three of those 8 are equally eligible to be called “the first three books for $A$”, so we count this one case at least $C(8.3) = 56$ times. In fact any three of $B$’s books is also eligible to be “the first three”, so we’ve counted this case at least $56*20 = 1120$ times! – Erick Wong Apr 15 at 17:24
• @lulu Yes, all books are to be distributed. – Lafinur Apr 15 at 17:52
• @ErickWong, you do make a point. I'll keep trying to work this out. – Lafinur Apr 15 at 17:53

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,

$$n=16172.$$

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.

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

• You are right, it was ambigious. I edited it. Unfortunately, the case was $14$ different books, so this is not the solution. I apologize and thank you for your effort anyway. – Lafinur Apr 17 at 16:49