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Lesson 3 of 6

Permutations and Combinations · Lesson 3 of 6

Repeated Objects and Restricted Arrangements

“Adjust arrangement counts for identical objects, fixed positions and objects that must stay together.”

Learning Objectives

• Derive why repeated identical objects require division by factorials. • Count arrangements with one or several repeated kinds. • Handle leading-zero and even-number restrictions. • Treat adjacent objects as a block and count its internal arrangements. • Use complements accurately for “not all together” conditions. • Check permissible values when solving permutation equations.

Identical objects change what counts as different

The word ROOT has four letters, but the two Os look identical. If we exchange only the Os, the visible word stays ROOT. The count 4! would be appropriate for four distinguishable labels; it overcounts arrangements of the letters we actually see.

Temporarily label the Os as O1 and O2. The labelled strings RO1O2T and RO2O1T become the same visible string ROOT when the labels are removed. Every visible arrangement corresponds to exactly 2! labelled arrangements. Therefore dividing 4! by 2! corrects the same overcount for every visible outcome.

Visible arrangementTwo labelled arrangements that produce it
ROOTRO1O2T; RO2O1T
ROTORO1TO2; RO2TO1
OORTO1O2RT; O2O1RT

The twelve distinct arrangements are ROOT, ROTO, RTOO, OROT, ORTO, OORT, OOTR, OTRO, OTOR, TROO, TORO and TOOR. Listing this small example verifies the division: 4!/2! = 12. We do not discard arrangements arbitrarily; we combine equally sized groups that become identical after labels disappear.

Several repeated kinds

In INSTITUTE, I occurs twice and T occurs three times. Temporarily distinguish both Is and all three Ts. For one visible arrangement, the Is can exchange their labels in 2! ways and the Ts can exchange theirs in 3! ways. These label exchanges can occur together, producing 2! × 3! labelled copies of each visible arrangement.

Arrangements with repeated kindsLaTeX
Arrange all n objects. Each pᵢ is the number of identical objects of one kind. Any remaining objects are distinct. The repeated-group counts must not exceed the total n.

For a single repeated kind with p copies, this becomes n!/p!. For INSTITUTE it gives 9!/(2!3!) = 30240. A letter occurring once would contribute 1! = 1, so we usually leave such factors out. The division applies to swapping indistinguishable copies, not to repeated positions or to arbitrary restrictions.

Example — ALLAHABAD

Problem
How many distinct arrangements use all the letters of ALLAHABAD?

  1. 1.Count the nine letters carefully: A appears four times, L twice, and H, B and D once each.
  2. 2.Temporarily distinguish all nine letters, giving 9! arrangements. Every visible arrangement has 4! label exchanges among the As and 2! among the Ls.
  3. 3.The required count is 9!/(4!2!) = 7560.
Example — Discs of three colours

Problem
Arrange 4 red, 3 yellow and 2 green discs in a row. Discs of the same colour are indistinguishable.

  1. 1.There are 4 + 3 + 2 = 9 positions and three repeated kinds.
  2. 2.Divide the labelled count 9! by the exchanges within each colour: 4!3!2!.
  3. 3.The count is 9!/(4!3!2!) = 1260. If same-colour discs carried different visible labels, this division would no longer describe the task.

Restrictions on individual positions

A valid number differs from a string of digits. Its first digit cannot be zero if it is to have the stated number of digits. It may also have a restricted final digit, such as an even units digit. Decide which positions are constrained, then count valid choices or remove clearly identified invalid arrangements.

Example — Four-digit numbers without zero

Problem
Use digits 1–9 to form four-digit numbers without repetition. How many are possible?

  1. 1.All nine digits can begin the number because zero is absent.
  2. 2.Fill the four positions in 9, 8, 7 and 6 ways.
  3. 3.The result is ⁹P₄=9×8×7×6=3024.
Example — A forbidden leading zero

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

  1. 1.These must be three-digit numbers. Direct counting gives 5 choices for the hundreds digit, then 5 remaining choices for the tens digit, then 4 for the units digit. Zero is allowed in the later positions.
  2. 2.The direct count is 5 × 5 × 4 = 100.
  3. 3.Alternatively count all ordered three-symbol strings: ⁶P₃=120. Those starting with zero have ⁵P₂=20 possible endings. Subtract them: 120 − 20 = 100.
  4. 4.The subtraction works because every invalid string has exactly the forbidden first digit, and every remaining string is a valid three-digit number.
Example — Even numbers without repetition

Problem
How many three-digit even numbers use digits 1, 2, 3, 4, 6 and 7 without repetition?

  1. 1.The units digit is restricted to 2, 4 or 6: 3 choices.
  2. 2.After fixing it, the hundreds digit has 5 choices and the tens digit has 4 choices. There is no zero to restrict the first position.
  3. 3.The count is 3 × 5 × 4 = 60. Choosing the units digit first makes the restriction easy to manage.

Objects that must stay together

If several objects must appear consecutively, temporarily bundle them as one block. Count arrangements of the block among the other objects, then multiply by the number of internal orders allowed inside the block. These two counts solve different parts of the arrangement.

A U EDGHTROne block + five consonants = six outer objectsOuter orders: 6! Internal vowel orders: 3! Total: 6! × 3!
A vowel block is one outer object— The three vowels occupy consecutive positions, but their internal order can change.
Example — Vowels together in DAUGHTER

Problem
Arrange all eight letters of DAUGHTER so that its vowels are together. Then count arrangements in which they are not all together.

  1. 1.The vowels A, U and E are distinct. Bundle them into one block. Together with the five distinct consonants, this gives six outer objects.
  2. 2.Arrange the outer objects in 6! ways. Inside the vowel block, the three vowels can be arranged in 3! ways. Multiply: 6! × 3! = 4320.
  3. 3.All eight letters without restriction can be arranged in 8! = 40320 ways. Subtract the all-together arrangements: 40320 − 4320 = 36000.
  4. 4.The subtraction counts arrangements where the three vowels do not form a single block. It permits two adjacent vowels if the third is elsewhere.
Common mistake

“Not all vowels together” is the complement of “all vowels together”. It differs from “no two vowels adjacent”. For example, an arrangement with AU adjacent and E separated belongs to the first condition but violates the second. Translate the wording before choosing a method.

Combining repetitions with restrictions

The block method also works with repeated letters, but both the outer objects and the internal block may contain identical copies. Keep separate inventories for them. Fixing a position removes exactly the letter placed there, so update the remaining multiplicities before counting.

Example — INDEPENDENCE and four restrictions

Problem
Count all arrangements of INDEPENDENCE, then those starting with P, those with all vowels together, those without all vowels together, and those beginning with I and ending with P.

  1. 1.There are twelve letters: N occurs 3 times, E occurs 4 times and D occurs 2 times; I, P and C occur once. Total arrangements: 12!/(3!4!2!) = 1663200.
  2. 2.Fix P first. It has no repeated copies, so the remaining eleven letters retain the same repeated counts. The number is 11!/(3!4!2!) = 138600.
  3. 3.The five vowels are E, E, E, E and I. Bundle them. The outer collection consists of the vowel block and seven consonants: eight objects with three Ns and two Ds. Its count is 8!/(3!2!).
  4. 4.Inside the block, the count is 5!/4! = 5. Multiply to get [8!/(3!2!)] × [5!/4!] = 16800.
  5. 5.Subtract this from the unrestricted total for “not all together”: 1663200 − 16800 = 1646400.
  6. 6.Fix I first and P last. Ten letters remain, with three Ns, four Es and two Ds. Their count is 10!/(3!4!2!) = 12600.

Solving equations without losing the meaning

Permutation notation sometimes appears in an equation with an unknown n or r. Expand the factorials or descending products, cancel only nonzero factors, and solve the resulting equation. Then return to the meaning of the original count: availability and selection counts must be whole numbers in the allowed range.

Example — An unknown number of objects

Problem
Find n > 4 if ⁿP₅=42 ⁿP₃.

  1. 1.Expand both counts: n(n − 1)(n − 2)(n − 3)(n − 4) = 42n(n − 1)(n − 2).
  2. 2.Since n > 4, the common factor n(n − 1)(n − 2) is nonzero. Cancel it to obtain (n − 3)(n − 4) = 42.
  3. 3.Expand and rearrange: n² − 7n − 30 = 0. Factor: (n − 10)(n + 3) = 0.
  4. 4.The roots are 10 and −3. A negative number of available objects is invalid, so n = 10.
Example — A ratio of permutations

Problem
Find n > 4 if ⁿP₄/⁽ⁿ⁻¹⁾P₄=5/3.

  1. 1.Write the numerator as n(n − 1)(n − 2)(n − 3) and the denominator as (n − 1)(n − 2)(n − 3)(n − 4).
  2. 2.The common factors are nonzero. Their cancellation gives n/(n − 4) = 5/3.
  3. 3.Cross-multiply: 3n = 5n − 20. Therefore 2n = 20 and n = 10, which satisfies n > 4.
Example — An algebraic root that must be rejected

Problem
Solve 5 ⁴Pᵣ=6 ⁵P₍ᵣ₋₁₎.

  1. 1.For both permutation counts to exist, r is a whole number and 1 ≤ r ≤ 4. Write this restriction before solving.
  2. 2.Using factorials gives 5 × 4!/(4 − r)! = 6 × 5!/(6 − r)!. Since 5 × 4! = 5!, cancel it.
  3. 3.Expand (6 − r)! = (6 − r)(5 − r)(4 − r)! and cancel the common factorial. This yields (6 − r)(5 − r) = 6.
  4. 4.Rearrange: r² − 11r + 24 = 0, or (r − 8)(r − 3) = 0.
  5. 5.The algebraic roots are 8 and 3, but 8 lies outside 1 ≤ r ≤ 4. Only r = 3 is valid. Check: 5 × ⁴P₃ = 5 × 24 = 120 and 6 × ⁵P₂ = 6 × 20 = 120.

Quiz

Quick check

How many distinct arrangements use all the letters of ROOT?

Quick check

How many four-digit numbers from 0–9 have no repeated digit?

Quick check

All vowels in DAUGHTER form one block. Which expression counts the arrangements?

Quick check

What does total arrangements minus all-vowels-together arrangements count?

Quick check

For ⁴Pᵣ, which algebraic root is invalid?

Quick check

If identical copies have counts 3 and 2 among seven objects, what is the full-arrangement count?

Practice Problems

Practice Problems
  1. Count four-digit numbers from 0–9 without repeated digits, explaining the first-position restriction.
  2. Count four-digit numbers from 1–5 without repetition, then count the even ones.
  3. Solve ⁽ⁿ⁻¹⁾P₃/ⁿP₄=1/9 and check the domain.
  4. Solve ⁵Pᵣ=2 ⁶P₍ᵣ₋₁₎ and ⁵Pᵣ=⁶P₍ᵣ₋₁₎, rejecting invalid roots.
  5. Count all arrangements of EQUATION, and arrangements of MONDAY using four letters, all letters, and all letters with a vowel first.
  6. Count arrangements of MISSISSIPPI in which its four Is are not all together. State whether your count permits adjacent pairs of Is.
  7. For PERMUTATIONS, count arrangements starting with P and ending with S, arrangements with all vowels together, and arrangements with exactly four letters between P and S.

Key Takeaways

Key Takeaways

• Swapping identical copies produces no new visible arrangement. • Divide n! by the factorial of each identical group when arranging all objects. • Fix restricted positions and update the remaining objects. • A block has an outer position and its own internal arrangements. • Complements require precise wording about the excluded event. • Reject algebraic roots that violate the original counting domain.

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Permutations of Distinct Objects

Next · Lesson 4

Choosing Without Regard to Order