Skip to main content
Engineering LibreTexts

14.4: The Division Rule

  • Page ID
    48396
    • Eric Lehman, F. Thomson Leighton, & Alberty R. Meyer
    • Google and Massachusetts Institute of Technology via MIT OpenCourseWare
    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Counting ears and dividing by two is a silly way to count the number of people in a room, but this approach is representative of a powerful counting principle.

    A \(k\)-to-1 function maps exactly \(k\) elements of the domain to every element of the codomain. For example, the function mapping each ear to its owner is 2-to-1. Similarly, the function mapping each finger to its owner is 10-to-1, and the function mapping each finger and toe to its owner is 20-to-1. The general rule is:

    Rule 14.4.1 (Division Rule). If \(f : A \rightarrow B\) is \(k\)-to-1, then \(|A| = k \cdot |B|\).

    For example, suppose \(A\) is the set of ears in the room and \(B\) is the set of people. There is a 2-to-1 mapping from ears to people, so by the Division Rule,\(|A| = 2 \cdot |B|\). Equivalently, \(|B| = |A| / 2\), expressing what we knew all along: the number of people is half the number of ears. Unlikely as it may seem, many counting problems are made much easier by initially counting every item multiple times and then correcting the answer using the Division Rule. Let’s look at some examples.

    Another Chess Problem

    In how many different ways can you place two identical rooks on a chessboard so that they do not share a row or column? A valid configuration is shown in Figure 14.2(a), and an invalid configuration is shown in Figure 14.2(b).

    clipboard_edc4b36a20bef909fe3d9716fe2ecf279.png
    Figure 14.2 Two ways to place 2 rooks on a chessboard. The configuration in (b) is invalid because the rooks are in the same column.

    Let A be the set of all sequences

    \[\nonumber (r_1, c_1, r_2, c_2)\]

    where \(r_1\) and \(r_2\) are distinct rows and \(c_1\) and \(c_2\) are distinct columns. Let \(B\) be the set of all valid rook configurations. There is a natural function \(f\) from set \(A\) to set \(B\); in particular, \(f\) maps the sequence \((r_1, c_1, r_2, c_2)\) to a configuration with one rook in row \(r_1\), column \(c_1\) and the other rook in row \(r_2\), column \(c_2\).

    But now there’s a snag. Consider the sequences:

    \[\nonumber (1, a, 8, h) \quad \text{and} \quad (8, h, 1, a)\]

    The first sequence maps to a configuration with a rook in the lower-left corner and a rook in the upper-right corner. The second sequence maps to a configuration with a rook in the upper-right corner and a rook in the lower-left corner. The problem is that those are two different ways of describing the same configuration! In fact, this arrangement is shown in Figure 14.2(a).

    More generally, the function \(f\) maps exactly two sequences to every board configuration; \(f\) is a 2-to-1 function. Thus, by the quotient rule, \(|A| = 2 \cdot |B|\). Rearranging terms gives:

    \[\nonumber |B| = \dfrac{|A|}{2} = \dfrac{(8 \cdot 7)^2}{2}.\]

    On the second line, we’ve computed the size of \(A\) using the General Product Rule just as in the earlier chess problem.

    Knights of the Round Table

    In how many ways can King Arthur arrange to seat his \(n\) different knights at his round table? A seating defines who sits where. Two seatings are considered to be the same arrangement if each knight sits between the same two knights in both seatings. An equivalent way to say this is that two seatings yield the same arrangement when they yield the same sequence of knights starting at knight number 1 and going clockwise around the table. For example, the following two seatings determine the same arrangement:

    clipboard_eccad1e17e758b13a4ef909f6708f53b9.png

    A seating is determined by the sequence of knights going clockwise around the table starting at the top seat. So seatings correspond to permutations of the knights, and there are \(n!\) of them. For example,

    clipboard_e17fa51f731b06809935033022ae5731d.png

    Two seatings determine the same arrangement if they are the same when the table is rotated so knight 1 is at the top seat. For example with \(n = 4\), there are 4 different sequences that correspond to the seating arrangement:

    clipboard_e75ebdf623d50cace3a1cd3342e59d852.png

    This mapping from seating to arrangments is actually an \(n\)-to-1 function, since all \(n\) cyclic shifts of the sequence of knights in the seating map to the same arrangement. Therefore, by the division rule, the number of circular seating arrangements is:

    \[\nonumber \dfrac{\text{# seatings}}{n} = \dfrac{n!}{n} = (n-1)!.\]


    This page titled 14.4: The Division Rule is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Eric Lehman, F. Thomson Leighton, & Alberty R. Meyer (MIT OpenCourseWare) .

    • Was this article helpful?