What this chapter is about
This chapter introduces systematic methods for counting the number of ways to arrange or select objects from a given set. When you need to find how many different seating arrangements are possible for a group of people, or how many ways you can choose a committee from a class, you are dealing with counting problems that become tedious if done by listing every possibility. Permutations and combinations provide formulas that handle such problems efficiently.
A permutation counts arrangements where order matters — arranging books on a shelf, forming different numbers from given digits, or scheduling events in a sequence. A combination counts selections where order does not matter — choosing team members, selecting questions from a paper, or picking fruits from a basket. The chapter builds on the fundamental principle of counting and extends to situations involving repetition and restrictions.
After studying this chapter, you should be able to distinguish between problems requiring permutations and those requiring combinations, apply the correct formula, and solve multi-step counting problems involving conditions like certain objects being together or certain positions being fixed.
Key ideas
- Fundamental Principle of Counting: If one event can occur in m ways and, after it occurs, a second event can occur in n ways, then the two events together can occur in m × n ways. This extends to any number of successive events.
- Factorial notation: n! (read as n factorial) equals n × (n−1) × (n−2) × ... × 3 × 2 × 1 for any positive integer n. By definition, 0! = 1.
- Permutation: A permutation is an arrangement of objects in a definite order. The number of permutations of n distinct objects taken r at a time is denoted P(n, r) or ⁿPᵣ.
- Combination: A combination is a selection of objects where order does not matter. The number of combinations of n distinct objects taken r at a time is denoted C(n, r) or ⁿCᵣ or (n choose r).
- Relation between permutations and combinations: P(n, r) = C(n, r) × r!, because each combination of r objects can be arranged in r! ways to give permutations.
- Permutations with repetition: When certain objects are identical, divide by the factorials of their repetition counts to avoid overcounting.
- Combinations have symmetry: C(n, r) = C(n, n−r), which means choosing r objects is equivalent to leaving out n−r objects.
Formulas and facts to remember
- Factorial: n! = n × (n−1) × (n−2) × ... × 1; also 0! = 1 and 1! = 1.
- Permutation formula: P(n, r) = n! / (n−r)!, valid for 0 ≤ r ≤ n.
- Combination formula: C(n, r) = n! / [r! × (n−r)!], valid for 0 ≤ r ≤ n.
- Permutations of n objects where p are alike, q are alike, etc.: Number of arrangements = n! / (p! × q! × ...).
- Sum property: C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n) = 2ⁿ.
- Pascal's identity: C(n, r) + C(n, r−1) = C(n+1, r).
- Special values: P(n, n) = n!, P(n, 1) = n, C(n, 0) = 1, C(n, n) = 1.
Worked examples
### Example 1: Forming three-digit numbers
Problem: How many three-digit numbers can be formed using the digits 2, 4, 6, 8 if no digit is repeated?
Solution: We have 4 digits and need to fill 3 places.
Hundreds place: 4 choices (any of the four digits). Tens place: 3 choices (one digit is used). Units place: 2 choices (two digits are used).
By the fundamental principle: 4 × 3 × 2 = 24.
Alternatively, this is P(4, 3) = 4! / (4−3)! = 4! / 1! = 24.
Answer: 24 three-digit numbers can be formed.
---
### Example 2: Selecting a committee
Problem: A class has 9 boys and 6 girls. In how many ways can a committee of 4 members be formed if it must contain at least 2 girls?
Solution: The committee must have at least 2 girls, so the possible cases are:
Case 1: Exactly 2 girls and 2 boys Ways = C(6, 2) × C(9, 2) = 15 × 36 = 540.
Case 2: Exactly 3 girls and 1 boy Ways = C(6, 3) × C(9, 1) = 20 × 9 = 180.
Case 3: Exactly 4 girls and 0 boys Ways = C(6, 4) × C(9, 0) = 15 × 1 = 15.
Total ways = 540 + 180 + 15 = 735.
Answer: 735 committees can be formed.
---
### Example 3: Arranging letters with repetition
Problem: How many distinct arrangements can be made using all the letters of the word BANANA?
Solution: The word BANANA has 6 letters: B appears 1 time, A appears 3 times, N appears 2 times.
For permutations with identical objects: Number of arrangements = 6! / (1! × 3! × 2!) = 720 / (1 × 6 × 2) = 720 / 12 = 60.
Answer: 60 distinct arrangements are possible.
Common mistakes
- Confusing permutations with combinations → Ask yourself: does the order of selection matter? If yes, use permutation; if no, use combination.
- Forgetting that 0! = 1 → This is a definition, not a calculation; accept it and use it in formulas.
- Using P(n, r) when objects repeat → When identical objects exist, use n! / (p! × q! × ...) instead of the standard permutation formula.
- Miscounting in at least or at most problems → Break into mutually exclusive cases, calculate each, then add.
- Applying C(n, r) with r greater than n → This is undefined; always verify that r ≤ n before using the formula.
Quick revision
- Permutation = arrangement (order matters); Combination = selection (order does not matter).
- P(n, r) = n! / (n−r)!; C(n, r) = n! / [r! × (n−r)!].
- For repeated objects in a word or set, divide n! by the factorials of the repetition counts.
- C(n, r) = C(n, n−r) — choosing r is the same as leaving n−r.
- Always identify whether the problem involves arrangement or selection before choosing a formula.