Further Permutations and Combinations

STA 211 · Lecture 01

Introduction

Many elementary probability calculations reduce to counting. When all outcomes in a finite sample space are equally likely,

$$P(A) = \frac{\text{number of outcomes favourable to }A}{\text{total number of outcomes}}.$$

The difficulty is usually not the final division but the accurate counting of outcomes. Permutations are used when order matters; combinations are used when order does not matter.

Learning Objectives

By the end of this lecture, you should be able to:

• apply the fundamental counting principle;
• distinguish permutations from combinations;
• derive and use $n!$, $_{n}P_{r}$ and $\binom{n}{r}$; and
• solve counting problems with restrictions.

Pre-Test

1. How many three-digit codes can be formed from the digits 0 to 9 if repetition is allowed?
2. Does the selection of three committee members depend on the order in which they are named?

The Fundamental Counting Principle

If a procedure has $k$ successive stages and stage $i$ can be completed in $n_{i}$ ways, then the whole procedure can be completed in

$$N = n_{1}n_{2}\cdots n_{k}$$

ways, provided the number of choices at each stage is correctly determined after earlier choices have been made.

Example 1.1: Product choices

A cafeteria offers 4 main dishes, 3 side dishes and 2 drinks. The number of distinct meals containing one item from each category is

$$4 \times 3 \times 2 = 24.$$

Factorials and Permutations

For a positive integer $n$,

$$n! = n(n - 1)(n - 2)\cdots 2 \cdot 1,$$

with the convention $0! = 1$.

The number of ordered arrangements of $r$ objects selected without replacement from $n$ distinct objects is

$$_{n}P_{r} = n(n - 1)\cdots(n - r + 1) = \frac{n!}{(n - r)!}.$$

When all $n$ objects are arranged,

$$_{n}P_{n} = n!.$$

Example 1.2: Office holders

A committee of 10 members must choose a chairperson, secretary and treasurer. Since the offices are different, order matters:

$$_{10}P_{3} = 10 \times 9 \times 8 = 720.$$

Combinations

The number of unordered selections of $r$ objects from $n$ distinct objects is

$$\binom{n}{r} =_{n}C_{r} = \frac{n!}{r!(n - r)!},\quad\quad 0 \leq r \leq n.$$

Each combination of $r$ objects can be arranged in $r!$ orders, so

$$_{n}P_{r} = r!\binom{n}{r}.$$

Useful identities include

$$\binom{n}{r} = \binom{n}{n - r},$$

$$\binom{n}{0} = \binom{n}{n} = 1,$$

and Pascal’s identity,

$$\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}.$$

Example 1.3: Committee composition

A committee of 4 is selected from 7 lecturers and 5 postgraduate students.

(a) No restriction

$$\binom{12}{4} = 495.$$

(b) Exactly 2 lecturers and 2 students

$$\binom{7}{2}\binom{5}{2} = 21 \times 10 = 210.$$

Example 1.4: At least condition

From 6 men and 4 women, form a committee of 5 containing at least 3 women. The mutually exclusive possibilities are 3 women and 2 men, or 4 women and 1 man:

$$\binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1} = 4(15) + 1(6) = 66.$$

Restricted Arrangements

Restrictions often require a case split, a block method or complementary counting.

Example 1.5: Two books together

Five distinct books are arranged on a shelf, and two specified books must be adjacent. Treat the pair as one block. The block and the other three books give $4!$ arrangements; the two books within the block can be ordered in $2!$ ways:

$$4!\, 2! = 48.$$

Common Errors

• Using combinations when offices or positions are distinct.
• Treating the first digit of a number as though it may be zero.
• Adding counts for cases that overlap.
• Forgetting that choices change when sampling is without replacement.

Summary

Situation	Formula
Arrange all $n$ distinct objects	$n!$
Arrange $r$ from $n$ distinct objects	$_{n}P_{r} = \frac{n!}{(n - r)!}$
Select $r$ from $n$ distinct objects	$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

Post-Test

1. Evaluate $_{12}P_{4}$ and $\binom{12}{4}$.
2. How many 5-letter strings can be formed from 26 letters if repetition is not allowed?
3. A class contains 8 women and 7 men. How many committees of 5 contain exactly 3 women?
4. In how many ways can 7 people sit in a row if two specified people must not sit together?
5. A password contains 3 distinct letters followed by 2 distinct digits. How many passwords are possible?

Selected answers: 1. $11,880$ and $495$; 2. $_{26}P_{5}$; 3. $\binom{8}{3}\binom{7}{2} = 1,176$; 4. $7! - 2(6!) = 3,600$; 5. $_{26}P_{3}\,_{10}P_{2} = 1,404,000$.

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