Permutations Involving Elements that are Alike

STA 211 · Lecture 02

Introduction

When some objects are indistinguishable, the ordinary factorial formula over-counts arrangements. The correction is made by dividing by the factorial of the number of repetitions of each type.

Learning Objectives

You should be able to:

• count distinguishable arrangements with repeated elements;
• apply the multinomial coefficient and multinomial theorem;
• count circular permutations; and
• distinguish linear, circular and reflected arrangements.

Pre-Test

1. How many distinct arrangements are possible for the letters in LEVEL?
2. Why is the number of arrangements around a round table not $n!$?

Repeated Elements

Suppose $n$ objects consist of $n_{1}$ objects of type 1, $n_{2}$ of type 2, …, $n_{k}$ of type $k$, where

$$n_{1} + n_{2} + \cdots + n_{k} = n.$$

The number of distinguishable linear arrangements is

$$\frac{n!}{n_{1}!n_{2}!\cdots n_{k}!}.$$

Example 2.1: CALCULUS

The word CALCULUS has 8 letters, with C, L and U each repeated twice. Therefore,

$$\frac{8!}{2!2!2!} = 5,040.$$

Example 2.2: Binary response patterns

A questionnaire has 10 Yes/No questions. A response pattern containing exactly 6 Yes responses and 4 No responses can be arranged in

$$\frac{10!}{6!4!} = \binom{10}{6} = 210$$

ways.

The Multinomial Coefficient

The coefficient

$$\binom{n}{n_{1},n_{2},\ldots,n_{k}} = \frac{n!}{n_{1}!n_{2}!\cdots n_{k}!}$$

counts the number of ways to divide $n$ distinct positions into labelled groups of sizes $n_{1},n_{2},\ldots,n_{k}$.

The multinomial theorem is

$$\left( x_{1} + x_{2} + \cdots + x_{k} \right)^{n} = \sum_{n_{1} + \cdots + n_{k} = n}^{}\frac{n!}{n_{1}!\cdots n_{k}!}x_{1}^{n_{1}}\cdots x_{k}^{n_{k}}.$$

Example 2.3: Coefficient in an expansion

The coefficient of $x^{3}y^{2}z$ in $(x + y + z)^{6}$ is

$$\frac{6!}{3!2!1!} = 60.$$

Circular Permutations

For $n$ distinct objects arranged around a circle, rotations are regarded as identical. Fix one object and arrange the remaining $n - 1$ objects:

$$N = (n - 1)!.$$

If clockwise and anticlockwise arrangements are also regarded as identical, as in a necklace that may be turned over, then for $n > 2$ the count is usually

$$N = \frac{(n - 1)!}{2}.$$

This division by 2 is valid only when each reflected arrangement has a distinct mirror image and no additional symmetry changes the count.

Example 2.4: Round-table seating

Nine people sit around a round table. The number of distinct seatings is

$$(9 - 1)! = 8! = 40,320.$$

Example 2.5: Alternating seats

Four men and four women sit around a round table in alternating order. Arrange the four men around the circle in $(4 - 1)!$ ways. Then place the four women in the four gaps in $4!$ ways:

$$3!\, 4! = 144.$$

Summary

The central principle is to divide the factorial count by the number of internal rearrangements that do not create visibly different outcomes.

Post-Test

1. Find the number of distinct arrangements of STATISTICS.
2. Find the coefficient of $a^{2}b^{3}c$ in $(a + b + c)^{6}$.
3. In how many ways can 7 people sit around a round table?
4. In how many ways can 5 married couples sit around a round table if men and women alternate?
5. How many sequences of 12 outcomes contain 5 successes, 4 failures of type 1 and 3 failures of type 2?

Selected answers: 1. $\frac{10!}{3!3!2!} = 50,400$; 2. $60$; 3. $6! = 720$; 4. $4!5! = 2,880$; 5. $\frac{12!}{5!4!3!} = 27,720$.

Previous lecture

Course home

Next lecture