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.

  • 6
    $\begingroup$ 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 $\sum_{i=3}^{11}\binom {14}i=16172$ $\endgroup$ – lulu Apr 15 at 17:09
  • $\begingroup$ That number is indeed too big. You should get something less than $2^{14} = 16384$. You have counted many cases multiple times $\endgroup$ – Henry Apr 15 at 17:09
  • 3
    $\begingroup$ 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! $\endgroup$ – Erick Wong Apr 15 at 17:24
  • $\begingroup$ @lulu Yes, all books are to be distributed. $\endgroup$ – Lafinur Apr 15 at 17:52
  • $\begingroup$ @ErickWong, you do make a point. I'll keep trying to work this out. $\endgroup$ – 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,



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

  • $\begingroup$ 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. $\endgroup$ – Lafinur Apr 17 at 16:49

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