📐
📊
✖️
← Back to Dashboard
Font Size:

1. Introduction

Counting is one of the most basic yet powerful activities in mathematics. When the number of possible arrangements or selections is huge, direct enumeration is impossible, and we need systematic methods. This chapter develops the fundamental principles of counting, the factorial notation, and the two central tools of combinatorics: permutations (arrangements) and combinations (selections).

The chapter begins with the fundamental principle of counting in its addition and multiplication forms. It then introduces factorial notation and develops formulas for the number of permutations of distinct objects, permutations with repetition, and permutations when some objects are alike. It defines combinations, establishes the relationship between permutations and combinations, and presents the standard properties of combination numbers.

Permutations and combinations are the backbone of probability theory, which is studied later in this very class. They also appear in geometry, algebra, and computer science, for instance in counting paths, forming committees, and analysing algorithms. A clear understanding of when to multiply and when to add, and when order matters, is the essence of this chapter.

2. The Fundamental Principle of Counting

Multiplication Principle

If one event can occur in m ways and a second independent event can occur in n ways after the first, then the two events together can occur in m x n ways.

Example: If a person has 3 shirts and 2 trousers, the number of ways of choosing one shirt and one trouser is 3 x 2 = 6.

Addition Principle

If one event can occur in m ways and another event can occur in n ways, and the two events cannot occur simultaneously, then one of them can occur in m + n ways.

Example: If a library has 4 books on mathematics and 5 books on physics, the number of ways of choosing one book of either subject is 4 + 5 = 9.

3. Factorial Notation

The product of the first n natural numbers is denoted by n! and read as "n factorial":

n! = 1 x 2 x 3 x ... x (n - 1) x n

By convention, 0! = 1. Also note that n! = n x (n - 1)!.

Some values: 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040.

Permutation of n Distinct Objects

The number of permutations (arrangements) of n distinct objects taken r at a time is:

nPr = n!/(n - r)! = n(n - 1)(n - 2)...(n - r + 1)

In particular, the number of permutations of n distinct objects taken all at a time is nPn = n!.

Example: The number of ways of arranging 3 letters A, B, C taking all 3 is 3! = 6.

4. Permutations When All Objects Are Not Distinct

If a set of n objects contains n1 objects of one kind, n2 objects of another kind, and so on, where n1 + n2 + ... + nk = n, then the number of distinct permutations is:

n!/(n1! n2! ... nk!)

Example: The number of distinct permutations of the letters of the word "APPLE" is 5!/2! = 60, because the letter P occurs twice.

5. Permutations with Repetition

If repetition of objects is allowed, then the number of permutations of n distinct objects taken r at a time is n^r, because each of the r positions can be filled in n ways.

Example: The number of 3-digit numbers that can be formed using the digits 1, 2, 3, 4 with repetition allowed is 4^3 = 64.

6. Combinations

A combination is a selection of objects in which order does not matter. The number of combinations of n distinct objects taken r at a time is:

nCr = n!/(r!(n - r)!)

Relation Between Permutations and Combinations

nPr = nCr x r!

Each selection of r objects can be arranged among themselves in r! ways, giving r! times as many permutations as combinations.

Example: The number of ways of choosing a committee of 2 from 5 people is 5C2 = 5!/(2!3!) = 10.

Values of nCr

These properties make the calculation of combination values much easier.

7. Selection When Some Objects Are Alike

When selections involve identical objects, the counting is different. The number of ways of selecting r objects from n distinct objects is nCr. If identical objects are present, problems are typically approached by direct case analysis or by considering the number of objects taken from each type.

For example, to find the number of ways of selecting any number of objects (from 0 to n) from n distinct objects, the answer is 2^n, since each object can either be chosen or not.

8. Applications and Word Problems

Word problems on permutations and combinations require careful reading:

Example: The number of ways in which 5 boys can be seated in a row is 5! = 120. The number of ways in which 5 boys can be seated around a circular table is (5 - 1)! = 4! = 24, since rotations are considered identical in a circle.

Quick Revision Tables

Table 1: Key Formulas

Situation Formula Example
Permutations of n objects taken r nPr = n!/(n-r)! 5P3 = 60
Permutations of n objects all at once nPn = n! 5! = 120
Permutations with repetition n^r 4^3 = 64
Permutations with alike objects n!/(n1!n2!...nk!) APPLE: 5!/2! = 60
Combinations nCr = n!/(r!(n-r)!) 5C2 = 10

Table 2: Properties of nCr

Property Statement
Symmetry nCr = nC(n-r)
Boundary values nC0 = 1, nCn = 1
Single selection nC1 = n, nC(n-1) = n
Pascal's identity nCr + nC(r-1) = (n+1)Cr
Total subsets nC0 + nC1 + ... + nCn = 2^n

Mind Map

graph TD A["Permutations and Combinations"] --> B["Counting Principles"] A --> C["Factorial"] A --> D["Permutations"] A --> E["Combinations"] A --> F["Applications"] B --> B1["Multiplication: m x n"] B --> B2["Addition: m + n"] C --> C1["n! = n x (n-1)!"] C --> C2["0! = 1"] D --> D1["nPr = n!/(n-r)!"] D --> D2["Repetition: n^r"] D --> D3["Alike objects: n!/(n1!n2!...)"] E --> E1["nCr = n!/(r!(n-r)!)"] E --> E2["nPr = nCr x r!"] E --> E3["nCr + nC(r-1) = (n+1)Cr"] F --> F1["Order matters: permutation"] F --> F2["Order does not matter: combination"] F --> F3["Circular arrangement: (n-1)!"]

Important Diagrams (SVG)

Diagram 1: The Multiplication Principle Visualised

Multiplication Principle: 3 shirts x 2 trousers = 6 outfits Start 3 ways Shirt 1 Shirt 2 Shirt 3 T1 T2 T1 T2 Each shirt combines with each trouser: 3 x 2 = 6 branches Addition principle: Choose 1 of 4 maths books or 1 of 5 physics books gives 4 + 5 = 9 ways (mutually exclusive choices). Golden Rule: Multiply for sequential choices, add for alternative choices.

Diagram 2: Permutation versus Combination

Permutation (order matters) vs Combination (order ignored) Permutation of A, B, C ABC, ACB, BAC, BCA, CAB, CBA Total = 3! = 6 arrangements nPr = n!/(n-r)! 3P3 = 6 Each selection has r! internal arrangements Combination of A, B, C (choose 2) AB, AC, BC Total = 3C2 = 3 selections nCr = n!/(r!(n-r)!) 3C2 = 3 AB is the same as BA in a selection Relation between them nPr = nCr x r! Example: 5P3 = 5C3 x 3! = 10 x 6 = 60 Golden Rule: Use permutations when order matters, combinations when it does not.

Common Mistakes

  1. Using the multiplication principle when the addition principle is required. Use multiplication for a sequence of choices and addition for mutually exclusive alternatives.
  2. Confusing nPr with nCr. Remember that nPr counts arrangements (order matters) and nCr counts selections (order does not matter).
  3. Forgetting that 0! = 1. Many students write 0! = 0, which breaks the formulas nC0 = 1 and nPn = n!.
  4. Applying n!/(n1!n2!...) without verifying that n1 + n2 + ... = n, the total number of objects.
  5. In circular permutations, using n! instead of (n - 1)! for seating people around a circle where rotations are identical.
  6. Misapplying the repetition formula: permutations with repetition allowed of n objects taken r at a time is n^r, not nPr.
  7. Forgetting the property nCr = nC(n-r), which is the most useful shortcut for reducing computations.
  8. Cancelling factors incorrectly in nCr; always write out the factorial definition before simplifying.
  9. In word problems, failing to check whether the objects being counted are distinguishable, especially with repeated letters.

Exam Tips

  1. Read the question and first decide: does the order matter? Seating, digits, and arrangements are permutations; committees and selections are combinations.
  2. Memorise factorial values up to 7! (5040) to speed up calculations: 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720.
  3. Use nCr = nC(n-r) to reduce big numbers, e.g., 100C98 = 100C2 = 4950.
  4. For words with repeated letters, divide by the factorial of each repeated letter's count: "MISSISSIPPI" has 11!/(4!4!2!) arrangements.
  5. When a problem says "at least", break it into cases and add: e.g., at least 2 girls from 5 means 5C2 + 5C3 + 5C4 + 5C5.
  6. For circular arrangements of n distinct objects, use (n - 1)!.
  7. Practise verifying answers with small examples, such as checking that 5P2 = 5C2 x 2! = 20 = 10 x 2.

Conclusion

Permutations and combinations provide a systematic way to count arrangements and selections without listing them. The fundamental principles of counting give the logical basis, the factorial notation supplies the computational engine, and the formulas for nPr and nCr, together with the properties of combination numbers, make the counting process efficient. The distinction between order-sensitive arrangements and order-insensitive selections is the key conceptual hurdle, and once it is mastered, word problems become straightforward. These tools are essential for probability, and they find application across mathematics, computer science, and every field that deals with counting and selection.