Permutations and Combinations | ISC Class 11 Maths Notes

This note covers the fundamental principle of counting, factorial notation, permutations, repetition and alike objects, restricted arrangements, formation of numbers, circular permutations, combinations, combination identities, selections involving alike objects, and mixed selection and arrangement problems.

How does the fundamental principle of counting work?

Counting finds how many possible outcomes satisfy stated conditions. Listing every outcome can become lengthy. A counting principle replaces the full list with a calculation, provided each outcome is counted once and every required outcome is included.

What does the multiplication principle count?

Definition: Let m be the number of choices for a first event and n the number of choices for a second event after each first choice. The number of ordered pairs of choices is m × n.

The symbol × means multiplication. For a third event with p choices after each preceding pair, where p is its number of available choices, the total becomes m × n × p. Choose a fixed order of stages and count the available choices at each stage.

Mohan has three pants and two shirts. Choosing one of each gives 3 × 2 = 6 outfits. The first choice does not remove either shirt. In an arrangement without repetition, however, using an object removes it from the choices for later positions.

A branching diagram labels the pants P₁, P₂ and P₃, and the shirts S₁ and S₂. Each pants branch divides into two shirt branches, ending in six labelled pairs.

See Fig. 6.1 in your NCERT textbook

Worked example 1. Sabnam has two school bags, three tiffin boxes and two water bottles. How many choices contain one item of each kind?

Answer: Choose the bag in 2 ways, the tiffin box in 3 ways and the bottle in 2 ways. The multiplication principle gives 2 × 3 × 2 = 12 choices.

When should separate counts be added?

Disjoint cases are cases that have no outcome in common. Add their counts when they exhaust the permitted alternatives. For signals made from five different flags arranged in order, one below another on a vertical staff, using two, three, four or five flags gives separate cases.

The counts are 20, 60, 120 and 120 respectively, so there are 320 signals altogether. Multiply choices within each signal length; add across the different lengths. A signal cannot belong to two different length cases, which prevents double counting.

What does factorial notation mean?

Definition: For a positive integer n, n! means the product of all positive integers from 1 to n. The symbol ! is read as “factorial”. Also, 0! is defined to equal 1.

A positive integer is a whole number greater than zero. Here n is the number whose factorial is being calculated. Thus 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorial notation abbreviates a product, rather than an ordinary multiplication by the exclamation mark.

Property: The factorial recurrence

n! = n(n − 1)! for a positive integer n. The symbol − means subtraction, and adjacent factors inside an expression are multiplied. The recurrence follows by separating the largest factor from the product defining the factorial.

For example, 5! can be written as 5 × 4!, or as 5 × 4 × 3!. Stop expanding when the remaining factorial cancels with one in the denominator. The denominator is the expression below a fraction bar; the numerator is the expression above it.

Worked example 2. Evaluate 7!/5! and 12!/(10! × 2!). Here / denotes division and the brackets group the complete denominator.

Answer: 7!/5! = (7 × 6 × 5!)/5! = 42. Also, 12!/(10! × 2!) = (12 × 11)/(2 × 1) = 66.

Why must the factorial stay attached to its argument?

In (n − r)!, where r is another integer, subtraction takes place before the factorial is evaluated. It does not mean n! − r!. Similarly, factorial does not distribute over addition or subtraction. Calculate each factorial separately when the terms themselves are being added or subtracted.

For instance, 7! − 5! = 5040 − 120 = 4920. This differs from cancelling factorials in a fraction, because subtraction does not create a common multiplicative factor that can simply be removed.

The value 0! = 1 makes the later counting formulae work when every object is used or when nothing is selected. Factorials in these formulae have non-negative integer arguments, meaning whole numbers that are zero or positive.

How are permutations of distinct objects calculated?

Definition: A permutation is an arrangement of some or all objects in a definite order. Distinct objects are objects that can be told apart. Changing their order can produce a different permutation.

Let n be the number of available distinct objects and r the number of positions to fill. Write P(n, r) for the number of arrangements using r of them without repetition. This notation is equivalent to ⁿPᵣ.

Theorem: Permutations without repetition

P(n, r) = n!/(n − r)!, where n and r are integers with 0 ≤ r ≤ n. The symbol ≤ means “less than or equal to”. No object may occupy more than one position.

1. The first position has n possible occupants because every object is still available.
2. After that position is filled, the second has n − 1 choices. The third has n − 2 choices.
3. The last of the r positions has n − r + 1 choices. Multiply these successive numbers of choices.
4. Writing the product with factorials gives n!/(n − r)!, because the unused lower factors cancel.

For r = n, all objects are arranged, and P(n, n) = n!/0! = n!. For r = 0, there is one empty arrangement, meaning the arrangement containing no objects, so P(n, 0) = 1.

Worked example 3. How many three-letter words, with or without meaning, can be made from NUMBER without repeating a letter?

Answer: Its six letters are distinct. The three positions have 6, 5 and 4 choices. Therefore P(6, 3) = 6!/3! = 6 × 5 × 4 = 120 words.

Why do different offices make order matter?

Choosing a chairman and a vice-chairman from twelve people, with nobody holding both offices, gives P(12, 2) = 12 × 11 = 132 assignments. The offices identify different positions. Interchanging the same two people produces a different assignment.

Before substituting into a formula, describe what each position represents. This distinguishes an ordered assignment from a selection in which the same people form one group regardless of how their names are listed.

How do repetition and alike objects change a count?

Two different situations need different rules. Repetition allowed means an available type may be used again in another position. Alike objects are indistinguishable copies already present in a fixed collection. Swapping those copies does not create a new visible arrangement.

Theorem: Arrangements with repetition allowed

If each of r positions can independently receive any of n distinct choices, the count is nʳ. Here nʳ means n multiplied by itself r times. Choosing a letter for one position leaves all n choices available for the next.

Using the letters of ROSE to fill four positions gives 4⁴ = 256 words when repetition is allowed. Without repetition, the count is 4! = 24. The word “with or without meaning” means that every permitted letter arrangement counts, including those that are not dictionary words.

Theorem: Permutations of a fixed collection with alike objects

Suppose n is the total number of objects. Let k be the number of repeated kinds, and let p₁, p₂, …, pₖ be the numbers of identical copies of those kinds. The dots indicate that the same pattern continues through the kth kind.

The number of arrangements using all objects is n!/(p₁!p₂!…pₖ!). Any remaining objects are distinct. Temporarily labelling identical copies gives n! arrangements; removing those artificial labels makes p₁!p₂!…pₖ! labelled arrangements coincide.

Worked example 4. Find the number of different arrangements using all the letters of ALLAHABAD.

Answer: There are nine letters, with four copies of A and two copies of L. H, B and D each occur once. Divide out the rearrangements of identical copies: 9!/(4! × 2!) = 7560.

The fixed stock matters: every arrangement still contains four As and two Ls. This is different from allowing any letter to be reused freely. The denominator removes indistinguishable exchanges; it does not remove any required letter from the word.

How are numbers formed from given digits?

A digit is a symbol used to write a number. Its position matters: exchanging digits can change the number. Apply the multiplication principle to the positions, with particular care about zero, repeated digits and restrictions on the final digit.

Which position should be filled first?

An even number is an integer divisible by two. Its units digit, the rightmost digit, must be even. When a condition restricts one position, counting that position first can make the remaining choices easier to identify.

Worked example 5. How many two-digit even numbers can be formed using 1, 2, 3, 4 and 5 if repetition is allowed?

Answer: The units place has two choices, 2 or 4. The tens place has all five choices because repetition is allowed. Therefore the number is 2 × 5 = 10.

With the digits 1 to 9 and no repetition, a four-digit number has 9, 8, 7 and 6 choices successively. Hence there are P(9, 4) = 3024 such numbers. None of these digits creates a leading-zero problem.

How is a leading zero excluded?

A leading zero is zero written in the leftmost position of a proposed number. It does not make that number have the intended number of digits. For example, 092 represents a two-digit number, so it cannot be counted as a three-digit number.

Worked example 6. How many numbers between 100 and 1000 can be formed from 0, 1, 2, 3, 4 and 5 without repeating a digit?

Answer: First count all three-position arrangements: P(6, 3) = 120. Those beginning with zero have P(5, 2) = 20 arrangements of the other positions. Subtract them to obtain 120 − 20 = 100 valid numbers.

This is counting by subtraction: count a larger collection and remove exactly the invalid outcomes. Check that the excluded collection matches the restriction. A digit list containing repeated entries supplies only the stated number of copies unless further repetition is expressly allowed.

How are restrictions such as together or apart handled?

A restricted permutation is an arrangement that satisfies an additional condition, such as fixed end letters or specified objects staying together. Translate the condition into a counting operation before evaluating factorials.

How does the block method work?

A block is a group temporarily treated as one object because its members must stay adjacent. Count the arrangements of the block with the remaining objects, then multiply by the permitted arrangements inside it. Identical objects still need the appropriate factorial division.

Worked example 7. Using all eight letters of DAUGHTER, how many arrangements have all three vowels together? Vowels here are A, U and E; the other letters are consonants.

Answer: Treat the three vowels as one block. Together with the five consonants, this gives six distinct objects to arrange in 6! ways. The vowels have 3! internal arrangements. The total is 6! × 3! = 4320.

There are 8! unrestricted arrangements. Therefore the number in which the three vowels are not all together is 8! − 6! × 3! = 36000. This subtraction still allows two vowels to be adjacent.

How does the gap method prevent adjacency?

A gap is a place before, between or after already arranged objects. To keep specified objects pairwise apart, place at most one of them in each selected gap. “No two together” is a stronger condition than “not all together”.

Worked example 8. In how many ways can five distinct girls and three distinct boys sit in a row if no two boys sit together?

Answer: Arrange the girls in 5! ways. Their row creates six gaps, including both ends. Assign the three boys to different gaps in P(6, 3) ways. The total is 5! × P(6, 3) = 14400.

For fixed end positions, first place the required objects and arrange the remainder. With INDEPENDENCE beginning with I and ending with P, the remaining letters include three Ns, four Es and two Ds. The count is 10!/(3! × 4! × 2!) = 12600.

How are circular permutations counted?

A circular permutation arranges objects around a circle. Before counting, decide whether rotations and reversals represent different arrangements. A rotation moves every object the same distance around the circle while preserving the order of its neighbours.

Result: Rotations identified, reverse orders distinguished

For n distinct objects around an unlabelled circle, with n a positive integer, rotations are regarded as the same arrangement. If clockwise and anticlockwise orders are distinguished, the count is (n − 1)!.

1. Choose one particular object as a reference. Its identity is fixed throughout the count.
2. Hold that object in one position to remove the freedom to rotate the whole arrangement.
3. Arrange the other n − 1 distinct objects in order around it.
4. There are (n − 1)! such orders, and each circular arrangement is represented once.

Clockwise means the direction followed by the hands of a clock; anticlockwise means the opposite direction. Reading a circular order in reverse generally changes which object follows which. Rotation alone does not turn one reading direction into the other.

Result: Reverse circular orders also identified

If reversed orders count as the same, the number is (n − 1)!/2 for n distinct objects with n at least 3. Each order pairs with its reverse after rotations have already been identified.

The restriction that n is at least 3 matters: dividing by two is not the rule for a circle containing just one or two distinct objects. Likewise, these formulae assume all objects are distinct. Repeated identical objects can create different symmetries, so the same division cannot be applied automatically.

Note: If all n positions are individually labelled and must be treated as different, assigning n distinct objects to them gives n! arrangements. Use (n − 1)! only when rotating the entire arrangement leaves it unchanged for the purpose of the question.

Identify the equivalence rule, meaning the rule that decides when two drawings represent one arrangement, before choosing the formula. The appearance of a circle by itself is insufficient to determine whether a division by n, or a further division by two, is justified.

What is a combination, and how is its formula derived?

Definition: A combination is a selection in which order is unimportant. Let C(n, r), also written ⁿCᵣ, be the number of selections of r objects from n distinct objects without repetition.

Let X, Y and Z name three tennis players. Selecting X and Y gives the same two-person team as selecting Y and X. The available teams are XY, YZ and ZX, where each pair of letters names its two members.

Three separate ovals contain the labels XY, YZ and ZX. Each oval represents one pair selected from players X, Y and Z.

See Fig. 6.3 in your NCERT textbook

Theorem: The connection between permutations and combinations

P(n, r) = C(n, r) × r!. Each selection of r distinct objects can be arranged in r! orders. Arrangements from different selections cannot coincide, since they do not contain exactly the same objects.

1. Select r of the n distinct objects in C(n, r) ways.
2. For each selection, arrange its members in r! ways.
3. Multiply to obtain C(n, r) × r!, which counts the same arrangements as P(n, r).
4. Divide by r! to obtain C(n, r) = n!/[r!(n − r)!], valid for integers 0 ≤ r ≤ n.

How do selection and arrangement differ?

Dividing by r! corrects the number of orders of each selected group. It does not mean that some groups are discarded. A question asking for a team usually counts membership, whereas named offices distinguish the roles occupied by those members.

For the distinct objects A, B, C and D, choosing two gives AB, AC, AD, BC, BD and CD. There are six selections. Arranging each pair in its two possible orders gives twelve permutations, illustrating the same relationship directly.

Which identities simplify combinations?

An identity is an equality that holds for all values allowed by its conditions. Combination identities let us simplify expressions and solve equations without expanding large factorials. Keep the conditions on the selection size in view throughout.

Property: Boundary values and complementary selections

C(n, 0) = C(n, n) = 1. There is one way to select nothing and one way to select all n objects. Also, C(n, r) = C(n, n − r) for integers 0 ≤ r ≤ n.

A complementary selection consists of the objects left out of another selection. Selecting r objects determines exactly which n − r objects are rejected. This gives a pairing between the two types of selection and explains why their counts agree.

If x and y are integer selection sizes between zero and n, then C(n, x) = C(n, y) implies x = y or x + y = n. The word “or” gives two possibilities. Do not assume the sum condition until the equal-size possibility has been considered.

Worked example 9. If C(n, 9) = C(n, 8), find C(n, 17), where n is an integer at least 9.

Answer: The lower entries 9 and 8 are unequal, so they must sum to n. Thus n = 17, and C(n, 17) = C(17, 17) = 1.

Identity: Addition of neighbouring combinations

C(n, r − 1) + C(n, r) = C(n + 1, r) for integers 1 ≤ r ≤ n. It follows by using the factorial formula and a common denominator, meaning the same denominator for both fractions.

The two numerators over r!(n − r + 1)! are n!r and n!(n − r + 1). Their sum is n!(n + 1) = (n + 1)!. The resulting fraction is exactly the factorial expression for C(n + 1, r).

When solving a counting equation, reject values that make a factorial argument negative or violate a stated selection size. Algebraic solutions, meaning values that satisfy the manipulated equation, must also fit the original counting problem.

How are committees selected under conditions?

A committee here is a group selected by membership, without different offices assigned. Use combinations within each group of available people. Multiply for choices that must occur together, and add disjoint alternatives.

How is an exact composition counted?

Worked example 10. A group contains two men and three women, all distinct people. How many three-person committees can be selected? How many contain exactly one man and two women?

Answer: Without a restriction there are C(5, 3) = 10 committees. For the specified composition, choose the man in C(2, 1) ways and the women in C(3, 2) ways. The required count is C(2, 1) × C(3, 2) = 6.

Exactly specifies one permitted number. At least specifies a minimum, while at most specifies a maximum. The available people and the total committee size still limit which cases are possible.

How are “at least” cases organised?

Suppose a group consists of four girls and seven boys, and a team has five members. To include at least three girls, it must contain either three girls and two boys, or four girls and one boy. Five girls are unavailable.

For at least one boy and one girl, the possible numbers of girls are one, two, three and four. The corresponding counts are 140, 210, 84 and 7. Adding gives 441 teams. No team is counted in two cases because its number of girls is fixed.

A useful check is to read each product aloud as “choose these members and choose those members”. Read the sum as “this composition or that composition”. These words help connect the arithmetic to the actual membership condition.

How are selections involving alike objects counted?

The formula C(n, r) assumes the n available objects are distinct. If copies of a kind are indistinguishable, selecting one copy instead of another does not create a different selection. Count the permitted quantities of each kind instead.

What information identifies a selection?

Let k be the number of kinds. For each kind i, where i labels a kind from 1 to k, let aᵢ be the available number of copies and xᵢ the number selected. Every aᵢ and xᵢ is a non-negative integer.

For a selection of exactly r objects, the conditions are x₁ + x₂ + … + xₖ = r and 0 ≤ xᵢ ≤ aᵢ for every kind. A permitted list of quantities identifies one selection when copies within each kind are indistinguishable.

The upper limits matter because the stock is finite. Counting a list that demands more copies than are available would introduce a selection that cannot be made. Merely dividing C(n, r) by the factorials of the available stocks does not generally solve this problem.

Result: Selections when the total size is unrestricted

For one kind with aᵢ identical copies, the selected quantity can range from zero to aᵢ, giving aᵢ + 1 choices. If different kinds can be chosen independently, there are (a₁ + 1)(a₂ + 1)…(aₖ + 1) selections, including the empty selection.

If at least one object must be selected, subtract one for the empty selection. This product counts all possible total sizes. For a fixed size r, retain only the quantity lists whose entries add to r.

Objects sharing a colour need not be indistinguishable unless the problem says so. State the interpretation: individually distinguishable objects are selected using combinations, whereas identical copies are counted by their quantities. The same descriptive word can occur in either kind of question.

This also separates selection from permutation of alike things. A selection records the quantities chosen; an arrangement records their order as well. The factorial division used for arranging a fixed collection addresses order and must not be transferred to selection without justification.

How are selection and arrangement combined in one problem?

A mixed problem requires more than one counting operation. Some conditions decide which objects are used, while others decide their positions. Separate those decisions and ensure that every completed outcome has one account in the calculation.

When should selection come before arrangement?

Worked example 11. How many five-letter words, with or without meaning, containing three vowels and two consonants can be formed without repetition from INVOLUTE? Its vowels are I, O, U and E; its consonants are N, V, L and T.

Answer: Select three vowels in C(4, 3) = 4 ways and two consonants in C(4, 2) = 6 ways. There are 24 selected groups. Each contains five distinct letters, with 5! arrangements. The total is 24 × 5! = 2880 words.

The combination factors choose the letters. The factorial puts them in order. Stopping at 24 would count selections instead of words, while using every available letter in the final factorial would arrange more letters than the question permits.

How can the conditions be checked systematically?

1. Identify a complete outcome: a number, word, assignment, circular order or selected group.
2. Decide whether order matters and whether copies or repeated uses are permitted.
3. List positional restrictions and membership restrictions separately, retaining words such as “exactly”, “all” and “at least”.
4. Choose multiplication, disjoint cases, subtraction, blocks or gaps to enforce those restrictions.
5. Check the answer against the original conditions, including available stock, leading zero and any circular equivalence rule.

For words with repeated letters, selecting different compositions can leave different multiplicities to arrange. Count and arrange each permitted composition using its own repeated-letter denominator before adding the results. A single arrangement factor is justified only when it applies to every selection counted.

A clear solution explains what each factor counts. The numerical calculation then follows from the explanation. This also makes it possible to detect omitted internal block orders, duplicated cases or an unjustified division before the final answer is reached.

Glossary

• Multiplication principle — A rule multiplying the available choices at successive stages to count complete ordered outcomes.
• Factorial — The product of positive integers up to a given positive integer, with zero factorial defined as one.
• Permutation — An arrangement of some or all available objects in a definite order.
• Combination — A selection of objects for which changing their order does not produce another selection.
• Distinct objects — Objects that can be distinguished from one another when counting arrangements or selections.
• Indistinguishable objects — Identical copies whose exchange does not create a different visible arrangement or selection.
• Repetition allowed — A condition permitting the same available choice to be used in more than one position.
• Restricted permutation — An arrangement satisfying additional requirements about positions, adjacency or the objects permitted.
• Block method — Counting adjacent objects as a single unit, then accounting for permitted internal arrangements.
• Gap method — Separating specified objects by placing them in different spaces around already arranged objects.
• Circular permutation — An arrangement around a circle, counted according to stated rules about rotations and reversals.
• Disjoint cases — Separate possibilities that share no outcomes and can therefore be added without double counting.
• Complementary selection — The collection left unchosen when a specified selection is made from all available objects.
• Leading zero — Zero in the first written position, which does not contribute a digit to the number's length.

Common errors and misconceptions

• Misconception: Selecting a team and assigning different offices use the same formula. Correct: A team is determined by its members; different offices distinguish the positions occupied. Use combinations for the former and permutations for the latter.
• Misconception: Zero factorial equals zero. Correct: 0! = 1. This preserves the formulae for selecting everything, selecting nothing and arranging all available objects.
• Misconception: Repeated letters always mean that any letter may be reused freely. Correct: A fixed word supplies a fixed number of each letter. Divide by factorials of repeated copies when arranging all its letters.
• Misconception: Every arrangement of three available digits represents a three-digit number. Correct: An arrangement beginning with zero does not. Exclude it when counting numbers of the required length.
• Misconception: “Not all together” means “no two together”. Correct: The first condition permits some adjacency. The second forbids every adjacent pair among the specified objects.
• Misconception: Treating objects as one block finishes the calculation. Correct: Also count the permitted orders inside the block, with factorial division for any indistinguishable copies.
• Misconception: Every circular arrangement requires division by two. Correct: Divide by two only when reversed orders are identified, under the stated distinct-object conditions.
• Misconception: C(n, r) counts selections from any collection, including identical copies. Correct: It counts selections from n distinct objects. For alike copies, count permitted quantities of each kind.

Exam-style questions with model answers

Q1. Evaluate 7!/5!, where ! denotes factorial, showing the cancellation. [2 marks]

1. Expand the numerator only as far as the denominator: 7! = 7 × 6 × 5!.
2. Cancel the common non-zero factor 5!, giving 7!/5! = 7 × 6 = 42.

Q2. How many three-letter words, with or without meaning, can be formed from NUMBER without repeating a letter? [3 marks]

1. NUMBER contains six distinct letters. A word is an ordered arrangement, so exchanging letters can produce another outcome even if the same three letters are used.
2. The first position has six choices. Once it is filled, five letters remain for the second position, and four remain for the third.
3. Apply the multiplication principle to these successive choices: 6 × 5 × 4 = 120 possible words.

Q3. Find the number of numbers between 100 and 1000 formed from 0, 1, 2, 3, 4 and 5 without repeating a digit. [3 marks]

1. The required numbers have three digits. Initially count all ordered arrangements of three of the six digits: 6 × 5 × 4 = 120.
2. This includes invalid arrangements beginning with zero. Fixing zero first leaves five choices for the second position and four for the third, giving 20 invalid arrangements.
3. Subtract those arrangements from the unrestricted count: 120 − 20 = 100. The remaining numbers satisfy both the length condition and the restriction against repeated digits.

Q4. Using every letter of DAUGHTER exactly once, find the number of arrangements in which its three vowels A, U and E are all together, and the number in which they are not all together. Words need not have meaning. [5 marks]

1. All eight letters are distinct. For the first condition, treat the three vowels as a single block so that they occupy adjacent positions in every arrangement counted.
2. The block and five consonants form six distinct objects. These objects can be arranged in 6! ways among the positions of the word.
3. Within each block position, the three distinct vowels have 3! orders. Therefore the number with all vowels together is 6! × 3! = 4320.
4. Without the adjacency restriction, the eight distinct letters have 8! arrangements. The cases with all vowels together form a subset of these arrangements.
5. Subtract that subset: 8! − 4320 = 36000 arrangements have the vowels not all together. This condition still permits two vowels to be adjacent.

Q5. A group has four distinct girls and seven distinct boys. How many five-member teams contain at least three girls? No offices are assigned. [4 marks]

1. A team is an unordered selection. At least three girls allows either three girls and two boys or four girls and one boy; five girls are unavailable.
2. For three girls and two boys, the count is C(4, 3) × C(7, 2) = 4 × 21 = 84, where C counts unordered selections.
3. For four girls and one boy, the count is C(4, 4) × C(7, 1) = 1 × 7 = 7.
4. The two cases have different numbers of girls and do not overlap. Their sum gives 84 + 7 = 91 teams.

Q6. Using INVOLUTE without repeating a letter, how many five-letter words contain three vowels and two consonants? Its vowels are I, O, U and E, and its consonants are N, V, L and T. Words need not have meaning. [5 marks]

1. There are four available vowels and four available consonants, all distinct. First choose the letters that will be used, before counting their positions in the word.
2. Select three of the four vowels in C(4, 3) = 4 ways, where C denotes the number of unordered selections of the indicated size.
3. Select two of the four consonants in C(4, 2) = 6 ways. Each vowel selection can be paired with each consonant selection, giving 4 × 6 = 24 groups.
4. Each chosen group contains five distinct letters. Their positions matter in a word, so each group produces 5! = 120 different arrangements.
5. Multiply the number of groups by the number of orders per group: 24 × 120 = 2880 words. Different groups cannot produce the same word because their letters differ.

Q7. For n distinct objects around an unlabelled circle, where n is an integer at least 3, derive the count when rotations are identical but reverse orders are different. Then state the count when reverse orders are also identical. [4 marks]

1. Choose one particular object as a reference and hold it fixed. This removes the freedom to rotate the complete arrangement without changing its circular order.
2. The remaining n − 1 objects can be placed in any order around the reference object, giving (n − 1)! arrangements.
3. This count keeps clockwise and anticlockwise orders distinct, as required by the first condition. Fixing the reference object has removed rotations only.
4. If reverse orders are also identified, each arrangement pairs with its reverse. Since the objects are distinct and n is at least 3, divide by two to obtain (n − 1)!/2.

Key takeaways

• Multiply choices for successive stages, and add disjoint cases that together cover every permitted outcome.
• Factorial notation abbreviates descending products; cancel shared factorial factors instead of expanding large numbers unnecessarily.
• For n distinct objects, P(n, r) counts ordered arrangements of r objects without repetition.
• When every position can reuse any available choice, multiply the same number of choices at every stage.
• For a fixed collection containing identical copies, divide the unrestricted factorial by factorials of the repeated multiplicities.
• Use blocks for objects required together and separate gaps for specified objects that cannot be adjacent.
• For circular arrangements, state whether rotations and reversed orders are identified before choosing a counting formula.
• Combinations count unordered selections of distinct objects; each selection produces r! permutations when its r members are arranged.
• For identical copies, count permitted quantities of each kind and enforce the available stock and required total.
• In mixed problems, choose the required objects first and then count their permitted arrangements without duplicating outcomes.

Test yourself

Why is P(n, n) equal to n!?

Its factorial formula gives n!/0!. Since 0! = 1, arranging all n distinct objects gives n! permutations.

How many four-letter words can be made from ROSE if repetition is allowed?

Each of the four positions has four choices. The count is 4⁴ = 256 words, with or without meaning.

Why are arrangements of ALLAHABAD divided by 4! × 2!?

The four As are identical, as are the two Ls. Their internal exchanges do not create different visible words.

Why does subtracting arrangements with all vowels together not enforce complete vowel separation?

The remaining arrangements may still contain adjacent pairs of vowels. Only arrangements containing the entire vowel block have been removed.

What does C(n, r) = C(n, n − r) express?

Every selection of r objects determines exactly one complementary selection of the n − r objects left out.

If C(n, 9) = C(n, 8), what is n?

Since the selection sizes differ, their sum equals n. Therefore n = 9 + 8 = 17.

When is (n − 1)!/2 valid for circular arrangements?

It applies to n distinct objects with n at least 3 when rotations and reversed circular orders both represent the same arrangement.

For identical copies of several kinds, what determines a fixed-size selection?

The quantity chosen from each kind determines it. Each quantity must respect its available stock, and the quantities must sum to the required size.