Permutations and Combinations · Lesson 6 of 6
Chapter Summary and Practice
“Connect all counting methods, justify their conditions and practise choosing a strategy for mixed problems.”
• Connect staged counting, permutations and combinations. • Select a formula by checking order, repetition and object identity. • Combine fixed positions, blocks, gaps and separate cases accurately. • Use factorial identities and domain checks to verify results. • Solve mixed problems and explain why each outcome is counted once.
One chapter, one underlying question
Throughout this chapter, the aim has been to count complete outcomes without listing them all. The formulas describe different kinds of outcomes, rather than competing shortcuts. A successful solution begins by explaining what the outcome is and why two outcomes count as different.
For an outfit, an outcome combines choices from different groups. For a code, it records symbols in ordered positions. For a committee, it records members without an internal order. For ROOT, exchanging identical Os does not change the visible result. The count must reflect these distinctions before any factorial is evaluated.
A chapter-wide method guide
Use this table as a check after interpreting the task. Each row states both a method and its conditions. A formula can be numerically evaluated yet still count the wrong outcomes if those conditions do not describe the problem.
| Situation | Count or method | Conditions and meaning |
|---|---|---|
| Successive required choices | Multiply stage counts | Each path has the stated number of choices at its next stage. |
| Ordered positions; repetition allowed | nʳ | n symbols remain available at each of r positions. |
| r ordered objects from n distinct objects | nPr = n!/(n − r)! | No repetition; 0 ≤ r ≤ n. |
| All n distinct objects arranged | n! | Every object is used once. |
| All objects arranged with identical groups | n!/(p1!p2!⋯pk!) | Divide only for exchanges of indistinguishable copies. |
| r objects selected from n distinct objects | nCr = n!/[r!(n − r)!] | Membership matters; order does not. |
| Objects required together | Outer arrangements × internal arrangements | Treat the required collection as one block, then arrange its contents. |
| No two specified objects adjacent | Arrange separators; choose gaps; assign objects | Include both end gaps; use at most one object per selected gap. |
| Separate allowed cases | Add their counts | Cases must not overlap and must cover all allowed possibilities. |
| Forbidden outcomes easy to count | Total − forbidden | The excluded set must be exactly the complement of the requirement. |
| Specified dictionary position | Count prefix groups | Update the remaining inventory after fixing each letter. |
The same problem may fit more than one description. Counting nonzero first digits directly and subtracting leading-zero strings can agree. Choosing objects and arranging them can agree with a permutation formula. Agreement between two independently explained methods is a useful check, but neither method is justified by agreement alone.
The relationships behind the formulas
The multiplication principle supports both permutation formulas. Without repetition, position counts descend; with repetition, they remain n. Factorial notation compresses a descending product. Dividing a permutation count by r! then removes the internal orders of each selected r-object group.
For repeated objects, the division has a different purpose from the combination division. Dividing by p! removes hidden label exchanges among p identical copies in a full arrangement. Dividing by r! converts an ordered arrangement of r distinct objects into an unordered selection. Keep those reasons separate even though both involve factorial denominators.
Work through mixed decisions
The following examples bring several chapter ideas together. Read the reasoning before the arithmetic. For each solution, identify why stages multiply, why alternatives add, and which conditions ensure that no outcome is counted twice.
Problem
Arrange every letter of EQUATION so that all vowels are together and all consonants are together.
- 1.The eight letters are distinct. The five vowels are E, U, A, I and O; the three consonants are Q, T and N.
- 2.There are two outer objects: a vowel block and a consonant block. Their order can be chosen in 2! ways.
- 3.Inside the vowel block there are 5! orders; inside the consonant block there are 3! orders.
- 4.Multiply: 2! × 5! × 3! = 2 × 120 × 6 = 1440. Counting just a vowel block among separate consonants would allow consonants on both sides, violating the second condition.
Problem
How many five-card selections from a standard deck contain exactly one king?
- 1.Choose the king from the 4 kings in ⁴C₁=4 ways.
- 2.The other four cards must be non-kings. There are 48 non-kings, giving ⁴⁸C₄=194580 selections.
- 3.Multiply the required choices: 4 × 194580 = 778320.
- 4.Choosing the other four from all 51 remaining cards would allow additional kings and would not enforce “exactly one”.
Problem
A collection contains 5 tasks in set A and 7 in set B. Choose 8 tasks altogether, including at least 3 from each set. How many selections are possible?
- 1.Let a be the number chosen from set A. Since at least three come from each set, a ≥ 3 and 8 − a ≥ 3. Also a ≤ 5 because set A contains five tasks.
- 2.The allowed splits (A, B) are (3, 5), (4, 4) and (5, 3).
- 3.Their counts are ⁵C₃ × ⁷C₅=10×21=210, ⁵C₄ × ⁷C₄=5×35=175, and ⁵C₅ × ⁷C₃=1×35=35.
- 4.Add the distinct compositions: 210 + 175 + 35 = 420. Every valid selection has exactly one of these splits.
Problem
Choose ten people from twenty-five. Three specified friends either all join or none join. How many groups are possible?
- 1.Separate cases according to whether the three friends are included. A group cannot belong to both cases.
- 2.If all three join, their places are fixed and seven more people are chosen from the other twenty-two: ²²C₇=170544.
- 3.If none join, all ten are chosen from those twenty-two: ²²C₁₀=646646.
- 4.Add: 170544 + 646646 = 817190. Choosing ten from all twenty-five and merely noting the friendship condition would not enforce it.
Problem
Use every digit 0, 1, 3, 5, 7 and 9 exactly once to form a six-digit number divisible by ten.
- 1.Divisibility by ten fixes the units digit as 0.
- 2.The five remaining nonzero digits fill the first five positions in 5! ways. Because zero is already last, no leading-zero arrangement remains.
- 3.The answer is 5! = 120. Fix the forced position before arranging the others.
Problem
Arrange all letters of ASSASSINATION with all Ss together.
- 1.There are thirteen letters: A occurs three times, S four times, I twice, N twice, and T and O once.
- 2.Bundle SSSS into one block. The outer collection has ten objects: the block, three As, two Is, two Ns, T and O.
- 3.The outer count is 10!/(3!2!2!) = 151200. Inside the block, 4!/4! = 1 because all four Ss are identical.
- 4.Thus the final count is 151200. An extra 4! factor would count exchanges of identical Ss as though they produced new words.
Check meaning as well as arithmetic
A counting answer should be a non-negative whole number. But that alone is not enough. Return to the task and check whether every counted result is valid, whether every valid result is included, and whether any result has been counted more than once.
For leading-zero restrictions, distinguish codes from numbers. For a repeated-letter inventory, use a letter only as many times as supplied. For a block, check both the number of outer objects and the internal permutations. For a committee, avoid an unnecessary arrangement factor. For an equation, check whole-number domains after solving; the root r = 8 cannot satisfy a count that chooses at most four objects.
“Not all together” and “no two adjacent” describe different sets. “At least three” includes every feasible count from three upward; “exactly three” includes only one count. Small changes in wording can change the complete counting model.
A brief history of counting ideas
The source traces counting arrangements and selections across several mathematical traditions. The history gives context to the chapter: people needed systematic counting long before modern notation made the formulas compact.
The chapter connects early Indian counting ideas with Sushruta’s combinations of tastes, Pingala’s study of verse patterns, and Jain work under the name Vikalpa. It discusses Mahavira’s general formulas and Bhaskaracharya’s treatment in Lilavati under Anka Pasha. It also notes Chinese, Greek and Latin contributions; Arabic and Hebrew astronomical applications; work associated with Rabbi ben Ezra and Levi ben Gerson; and Jacob Bernoulli’s Ars Conjectandi, published in 1713 after his death. These examples show the range of questions that led people to study arrangements and selections.
Mixed practice across the chapter
Use the quizzes to check your understanding of the counting model. Then solve the mixed problems with a short explanation of each stage or case. Try to justify your method before calculating large factorials, and use a second method where it offers a useful check.
Quiz
Four-position codes use digits 0–9 with repetition allowed. How many are there?
Six distinct people are available. How do the counts for a two-person committee and two different offices compare?
Which count gives arrangements of A, A, B, B, C?
Four distinct separator people create five gaps. Two distinct other people must not sit adjacent. What is the count?
A five-card hand has exactly one ace. Where should its other four cards be selected from?
Which identity follows by including or excluding one distinguished object?
A committee contains at least three girls, but only four girls are available. Which girl counts must be considered?
An arrangement is not an all-vowel block. Must every pair of vowels be separated?
Practice Problems
- Form five-letter strings using 2 vowels and 3 consonants chosen from DAUGHTER without repetition. Show both the selection and arrangement stages.
- Arrange EQUATION with its vowels together and consonants together. Explain why there are two outer block orders.
- Choose a seven-person committee from 9 boys and 4 girls with exactly 3 girls, at least 3 girls, and at most 3 girls. List the feasible cases for each condition.
- Arrange the eleven letters A, A, E, I, I, M, N, N, O, T, X in dictionary order. How many arrangements appear before the first beginning with E?
- Count six-digit numbers divisible by ten using 0, 1, 3, 5, 7 and 9 exactly once.
- Form four-letter strings with two different vowels and two different consonants from the English alphabet, which has 5 vowels and 21 consonants.
- Choose 8 tasks from sets of sizes 5 and 7, taking at least 3 from each. Show that your cases are complete.
- Count five-card selections containing exactly one king. Explain why a second king is forbidden among the remaining choices.
- Seat 5 distinct men and 4 distinct women in a row, with the women occupying the even-numbered positions.
- Choose ten people from twenty-five when three specified friends must either all join or all stay out.
- Arrange ASSASSINATION with all Ss together. Count the remaining repeated kinds after forming the block.
- Solve 5 × ⁴Pᵣ = 6 × ⁵P₍ᵣ₋₁₎. Explain why only one of the quadratic roots is permitted.
Key Takeaways
• Start with the meaning of an outcome, then decide whether order matters. • Repetition allowed and identical copies are different counting conditions. • Every factor should represent a required choice; every added term should represent a separate case. • Blocks enforce togetherness, while distinct occupied gaps enforce separation. • Use symmetry, alternative methods and small lists to check the model. • A complete solution counts each valid outcome exactly once and respects every domain restriction.
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Combining Counting Methods
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