Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

3.1: The Idea of Distribution

Many of the problems we solved in Chapter 1 may be thought of as problems of distributing objects (such as pieces of fruit or ping-pong balls) to recipients (such as children). Some of the ways of viewing counting problems as distribution problems are somewhat indirect. For example, in Problem 37 you probably noticed that the number of ways to pass out \(k\) ping-pong balls to \(n\) children so that no child gets more than one is the number of ways that we may choose a \(k\)-element subset of an \(n\)-element set. We think of the children as recipients and objects we are distributing as the identical ping-pong balls, distributed so that each recipient gets at most one ball. Those children who receive an object are in our set. It is helpful to have more than one way to think of solutions to problems. In the case of distribution problems, another popular model for distributions is to think of putting balls in boxes rather than distributing objects to recipients. Passing out identical objects is modeled by putting identical balls into boxes. Passing out distinct objects is modeled by putting distinct balls into boxes.

3.1.1 The Twenty-Fold Way

When we are passing out objects to recipients, we may think of the objects as being either identical or distinct. We may also think of the recipients as being either identical (as in the case of putting fruit into plastic bags in the grocery store) or distinct (as in the case of passing fruit out to children). We may restrict the distributions to those that give at least one object to each recipient, or those that give exactly one object to each recipient, or those that give at most one object to each recipient, or we may have no Table 3.1: An incomplete table of the number of ways to distribute \(k\) objects to \(n\) recipients, with restrictions on how the objects are received such restrictions. If the objects are distinct, it may be that the order in which the objects are received is relevant (think about putting books onto the shelves in a bookcase) or that the order in which the objects are received is irrelevant (think about dropping a handful of candy into a child’s trick or treat bag). If we ignore the possibility that the order in which objects are received matters, we have created \(2 \cdot 2 \cdot 4 = 16\) distribution problems. In the cases where a recipient can receive more than one distinct object, we also have four more problems when the order objects are received matters. Thus we have 20 possible distribution problems.

We describe these problems in Table 3.1. Since there are twenty possible distribution problems, we call the table the “Twenty-fold Way,” adapting terminology suggested by Joel Spencer for a more restricted class of distribution problems. In the first column of the table we state whether the objects are distinct (like people) or identical (like ping-pong balls) and then give any conditions on how the objects may be received. The conditions we consider are whether each recipient gets at most one object, whether each recipient gets at least one object, whether each recipient gets exactly one object, and whether the order in which the objects are received matters. In the second column we give the solution to the problem and the name of the mathematical model for this kind of distribution problem when the recipients are distinct, and in the third column we give the same information when the recipients are identical. We use question marks as the answers to problems we have not yet solved and models we have not yet studied. We give explicit answers to problems we solved in Chapter 1 and problems whose answers are immediate. The goal of this chapter is to develop methods that will allow us to fill in the table with formulas or at least quantities we know how to compute, and we will give a completed table at the end of the chapter. We will now justify the answers that are not question marks and replace some question marks with answers as we cover relevant material.

If we pass out \(k\) distinct objects (say pieces of fruit) to n distinct recipients (say children), we are saying for each object to which recipient it goes. Thus we are defining a function from the set of objects to the recipients. We saw the following theorem in Problem 13b.

Theorem 3 There are \(n^{k}\) functions from a \(k\)-element set to an \(n\)-element set.

We proved it in one way in Problem 13b and in another way in Problem 75. If we pass out \(k\) distinct objects (say pieces of fruit) to \(n\) indistinguishable recipients (say identical paper bags) then we are dividing the objects up into disjoint sets; that is, we are forming a partition of the objects into some number, certainly no more than the number \(k\) of objects, of parts. Later in this chapter (and again in the next chapter) we shall discuss how to compute the number of partitions of a \(k\)-element set into n parts. This explains the entries in row one of our table.

If we pass out \(k\) distinct objects to n recipients so that each gets at most one, we still determine a function, but the function must be one-to-one. The number of one-to-one functions from a \(k\)-element set to an \(n\) element set is the same as the number of one-to-one functions from the set \([k] = \{1, 2, . . . , k\}\) to an \(n\)-element set. In Problem 20 we proved the following theorem.

Theorem 4 If \(0 \leq k \leq n\), then the number of \(k\)-element permutations of an \(n\)-element set is

\[n^{\underline{k}} = n(n - 1) \dotsc (n-k+1) = n!/(n-k)!.\]

If \(k > n\) there are no one-to-one functions from a \(k\) element set to an \(n\) element set, so we define \(n^{\underline{k}}\) to be zero in this case. Notice that this is what the indicated product in the middle term of our formula gives us. If we are supposed to distribute k distinct objects to n identical recipients so that each gets at most one, we cannot do so if \(k > n\), so there are 0 ways to do so. On the other hand, if \(k \leq n\), then it doesn’t matter which recipient gets which object, so there is only one way to do so. This explains the entries in row two of our table.

If we distribute \(k\) distinct objects to n distinct recipients so that each recipient gets at least one, then we are counting functions again, but this time functions from a \(k\)-element set onto an \(n\)-element set. At present we do not know how to compute the number of such functions, but we will discuss how to do so later in this chapter and in the next chapter. If we distribute \(k\) identical objects to \(n\) recipients, we are again simply partitioning the objects, but the condition that each recipient gets at least one means that we are partitioning the objects into exactly \(n\) blocks. Again, we will discuss how to compute the number of ways of partitioning a set of \(k\) objects into \(n\) blocks later in this chapter. This explains the entries in row three of our table.

If we pass out \(k\) distinct objects to \(n\) recipients so that each gets exactly one, then \(k = n\) and the function that our distribution gives us is a bijection. The number of bijections from an \(n\)-element set to an \(n\)-element set is \(n!\) by Theorem 4. If we pass out \(k\) distinct objects to n identical recipients so that each gets exactly 1, then in this case it doesn’t matter which recipient gets which object, so the number of ways to do so is 1 if \(k = n\). If \(k \neq n\), then the number of such distributions is zero. This explains the entries in row four of our table.

We now jump to row eight of our table. We saw in Problem 37 that the number of ways to pass out \(k\) identical ping-pong balls to \(n\) children is simply the number of \(k\)-element subsets of an \(n\)-element set. In Problem 39d we proved the following theorem.

Theorem 5 If \(0 leq k leq n\), the number of \(k\)-element subsets of an \(n\)-element set is given by

\[\binom{n}{k} = \frac{n^{\underline{k}}}{k!} = \frac{n!}{k!(n-k)!}.\]

We define \(\binom{n}{k}\) to be 0 if \(k > n\), because then there are no \(k\)-element subsets of an \(n\)-element set. Notice that this is what the middle term of the formula in the theorem gives us. This explains the entries of row 8 of our table. For now we jump over row 9.

In row 10 of our table, if we are passing out \(k\) identical objects to \(n\) recipients so that each gets exactly one, it doesn’t matter whether the recipients are identical or not; there is only one way to pass out the objects if \(k = n\) and otherwise it is impossible to make the distribution, so there are no ways of distributing the objects. This explains the entries of row 10 of our table. Several other rows of our table can be computed using the methods of Chapter 1.

3.1.2 Ordered Functions

(\rightarrow \bullet\) 122. Suppose we wish to place \(k\) distinct books onto the shelves of a bookcase with \(n\) shelves. For simplicity, assume for now that all of the books would fit on any of the shelves. Also, let’s imagine that once we are done putting books on the shelves, we push the books on a shelf as far to the left as we can, so that we are only thinking about how the books sit relative to each other, not about the exact places where we put the books. Since the books are distinct, we can think of the first book, the second book and so on.

  1. How many places are there where we can place the first book?
  2. When we place the second book, if we decide to place it on the shelf that already has a book, does it matter if we place it to the left or right of the book that is already there?
  3. How many places are there where we can place the second book? Online hint. Since we have \(i − 1\) books placed, if we want to place book i on a shelf that already has some books, is sliding it in to the left of all the books already there different from placing it to the right of all the books already there or between two books already there?
  4. In how many ways may we place the ith book into the bookcase? Online hint.
  5. In how many ways may we place all the books?

123. Suppose we wish to place the books in Problem 122e (satisfying the assumptions we made there) so that each shelf gets at least one book. Now in how many ways may we place the books? Online hint.

The assignment of which books go to which shelves of a bookcase is simply a function from the books to the shelves. But a function doesn’t determine which book sits to the left of which others on the shelf, and this information is part of how the books are arranged on the shelves. In other words, the order in which the shelves receive their books matters. Our function must thus assign an ordered list of books to each shelf. We will call such a function an ordered function. More precisely, an ordered function from a set \(S\) to a set \(T\) is a function that assigns an (ordered) list of elements of \(S\) to some, but not necessarily all, elements of \(T\) in such a way that each element of \(S\) appears on one and only one of the lists.1 (Notice that although it is not the usual definition of a function from \(S\) to \(T\), a function can be described as an assignment of subsets of \(S\) to some, but not necessarily all, elements of \(T\) so that each element of \(S\) is in one and only one of these subsets.) Thus the number of ways to place the books into the bookcase is the entry in the middle column of row 5 of our table. If in addition we require each shelf to get at least one book, we are discussing the entry in the middle column of row 6 of our table. An ordered onto function is one which assigns a list to each element of \(T\).

In Problem 122e you showed that the number of ordered functions from a \(k\)-element set to an \(n\)-element set is \(\prod\limits^{k}_{i=1}(n+i-1)\). This product occurs frequently enough that it has a name; it is called the \(k\)th rising factorial power of n and is denoted by \(n^{\bar{k}}\). It is read as “\(n\) to the \(k\) rising.” (This notation is due to Don Knuth, who also suggested the notation for falling factorial powers.) We can summarize with a theorem that adds two more formulas for the number of ordered functions.

Theorem 6 The number of ordered functions from a \(k\)-element set to an \(n\)-element set is

\[n^{\bar{k}} = \prod\limits^{k}_{i=1}(n+i-1) = \frac{(n+k-1)!}{(n-1)!} = (n+k-1)^{\underline{k}}\].

Ordered functions explain the entries in the middle column of rows 5 and 6 of our table of distribution problems.

3.1.3 Multisets

In the middle column of row 7 of our table, we are asking for the number of ways to distribute \(k\) identical objects (say ping-pong balls) to \(n\) distinct recipients (say children).

\(\bullet\) 124. In how many ways may we distribute \(k\) identical books on the shelves of a bookcase with \(n\) shelves, assuming that any shelf can hold all the books? Online hint.

\(\bullet\) 125. A multiset chosen from a set S may be thought of as a subset with repeated elements allowed. To determine a multiset we must say how many times (including, perhaps, zero) each member of \(S\) appears in the multiset. The number of times an element appears is called its multiplicity. For example if we choose three identical red marbles, six identical blue marbles and four identical green marbles, from a bag of red, blue, green, white and yellow marbles then the multiplicity of a red marble in our multiset is three, while the multiplicity of a yellow marble is zero. The size of a multiset is sum of the multiplicities of its elements. For example if we choose three identical red marbles, six identical blue marbles and four identical green marbles, then the size of our multiset of marbles is 13. What is the number of multisets of size
\(k\) that can be chosen from an \(n\)-element set? Online hint.

\(\rightarrow\) 126. Your answer in the previous problem should be expressible as a binomial coefficient. Since a binomial coefficient counts subsets, find a bijection between subsets of something and multisets chosen from a set \(S\). Online hint.

127. How many solutions are there in nonnegative integers to the equation \(x_{1} + x_{2} + · · · + x_{m} = r\), where \(m\) and \(r\) are constants? Online hint.

128. In how many ways can we distribute \(k\) identical objects to \(n\) distinct recipients so that each recipient gets at least \(m\)? Online hint.

Multisets explain the entry in the middle column of row 7 of our table of distribution problems.

3.1.4 Compositions of Integers

\(\cdot\) 129. In how many ways may we put \(k\) identical books onto \(n\) shelves if each shelf must get at least one book? Online hint.

\(\cdot\) 130. A composition of the integer \(k\) into \(n\) parts is a list of \(n\) positive integers that add to \(k\). How many compositions are there of an integer \(k\) into \(n\) parts? Online hint.

\(\rightarrow\) 131. Your answer in Problem 130 can be expressed as a binomial coefficient. This means it should be possible to interpret a composition as a subset of some set. Find a bijection between compositions of \(k\) into \(n\) parts and certain subsets of some set. Explain explicitly how to get the composition from the subset and the subset from the composition. Online hint.

\(\cdot\) 130. A composition of the integer \(k\) into \(n\) parts is a list of \(n\) positive integers that add to \(k\). How many compositions are there of an integer \(k\) into \(n\) parts? Online hint.

\(\rightarrow\) 131. Your answer in Problem 130 can be expressed as a binomial coefficient. This means it should be possible to interpret a composition as a subset of some set. Find a bijection between compositions of \(k\) into \(n\) parts and certain subsets of some set. Explain explicitly how to get the composition from the subset and the subset from the composition. Online hint.

\(\cdot\) 132. Explain the connection between compositions of \(k\) into \(n\) parts and the problem of distributing k identical objects to \(n\) recipients so that each recipient gets at least one.

The sequence of problems you just completed should explain the entry in the middle column of row 9 of our table of distribution problems.

3.1.5 Broken Permutations and Lah Numbers

\(\rightarrow \cdot\) 133. In how many ways may we stack \(k\) distinct books into \(n\) identical boxes so that there is a stack in every box? Online hint.

We can think of stacking books into identical boxes as partitioning the books and then ordering the blocks of the partition. This turns out not to be a useful computational way of visualizing the problem because the number of ways to order the books in the various stacks depends on the sizes of the stacks and not just the number of stacks. However, instead of dividing a set up into non-overlapping parts, we may think of dividing a permutation (thought of as a list) of our \(k\) objects up into \(n\) ordered blocks. We will say that a set of ordered lists of elements of a set \(S\) is a broken permutation of \(S\) if each element of \(S\) is in one and only one of these lists. The number of broken permutations of a \(k\)-element set with n blocks is denoted by \(L(k, n)\). The number \(L(k, n)\) is called a Lah Number (this is standard) and, from our solution to Problem 133, is equal to \(k!\binom{k-1}{n-1}n!\).

The Lah numbers are the solution to the question “In how many ways may we distribute \(k\) distinct objects to \(n\) identical recipients if order matters and each recipient must get at least one?” Thus they give the entry in row 6 and column 3 of our table. The entry in row 5 and column 3 of our table will be the number of broken permutations with less than or equal to \(n\) parts. Thus it is a sum of Lah numbers.

We have seen that ordered functions and broken permutations explain the entries in rows 5 and 6 of our table.

In the next two sections we will give ways of computing the remaining entries.