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Overview

Arrangement problems ask us to place objects in order.

Common examples include:

  • Arranging letters to form words
  • Arranging digits to form numbers
  • Finding the rank of a word in dictionary order
  • Finding the sum of all numbers formed from a set of digits

These problems combine permutations, repeated objects, and positional thinking.

The main idea is to count arrangements systematically rather than listing them.

Arranging Distinct Objects

If nn objects are all distinct, then the number of arrangements is n!n!.

Example 1

How many ways are there to arrange the letters in MATHMATH?

The four letters are distinct.

So the number of arrangements is

4!=24.4!=24.

Arranging Objects With Repetition

If some objects are repeated, then arranging them as if they were distinct overcounts.

Suppose there are nn total objects, with repeated groups of sizes

a1,a2,…,ak.a_1,a_2,\ldots,a_k.

Then the number of distinct arrangements is

n!a1!a2!⋯ak!.\frac{n!}{a_1!a_2!\cdots a_k!}.

Example 2

How many distinct arrangements are there of the letters in LEVELLEVEL?

The word LEVELLEVEL has 55 letters.

The letter LL appears twice, and the letter EE appears twice.

So the number of distinct arrangements is

5!2!2!=30.\frac{5!}{2!2!}=30.

Restrictions in Arrangements

Many arrangement problems include restrictions.

Common restrictions include:

  • Certain objects must be together
  • Certain objects must not be together
  • Vowels and consonants must alternate
  • Digits cannot start with 00
  • Repeated objects are present

The usual strategy is to count the unrestricted arrangements first, then impose the restrictions carefully.

Objects Together

If several objects must stay together, treat them as a block.

Example 3

How many ways are there to arrange the letters A,B,C,D,EA,B,C,D,E if AA and BB must be together?

Treat AA and BB as one block.

Then we arrange the block together with C,D,EC,D,E, giving 44 objects:

4!.4!.

Inside the block, AA and BB can be arranged in

2!2!

ways.

So the total number of arrangements is

4!2!=48.4!2!=48.

Objects Not Together

If certain objects must not be together, it is often easier to use complement counting.

Example 4

How many ways are there to arrange the letters A,B,C,D,EA,B,C,D,E if AA and BB must not be together?

There are

5!5!

total arrangements.

From Example 3, the number where AA and BB are together is

4!2!.4!2!.

Therefore, the number where they are not together is

5!−4!2!=120−48=72.5!-4!2!=120-48=72.

Lexicographic Rank

The lexicographic rank of a word is its position when all arrangements are listed in dictionary order.

Usually, ranks start from 11.

To find the rank of a word:

  1. Look at the first position
  2. Count how many valid arrangements begin with a smaller possible letter
  3. Fix the actual first letter
  4. Repeat for the next position
  5. Add 11 for the word itself

Worked Rank Example

Example 5

Find the rank of MATHMATH among all arrangements of the letters A,H,M,TA,H,M,T in alphabetical order.

The letters in alphabetical order are

A,H,M,T.A,H,M,T.

We want the rank of

MATH.MATH.

First consider the first letter.

Letters smaller than MM are

AA

and

H.H.

If the first letter is AA, the remaining 33 letters can be arranged in

3!3!

ways.

If the first letter is HH, the remaining 33 letters can also be arranged in

3!3!

ways.

So the number of words before MATHMATH with smaller first letter is

2(3!)=12.2(3!)=12.

Now fix the first letter as MM.

The second letter is AA. There is no remaining letter smaller than AA, so no new words are counted here.

Now fix

MA.MA.

The remaining letters are

H,T.H,T.

The third letter in MATHMATH is TT. There is one smaller remaining letter, namely HH.

If the third letter were HH, the last letter would be forced. This gives

11

word before MATHMATH.

So the total number of words before MATHMATH is

12+1=13.12+1=13.

Therefore, the rank of MATHMATH is

14.14.

Rank With Repeated Letters

When letters repeat, the same method works, but each count must account for repetition.

Example 6

How many arrangements of LEVELLEVEL begin with a letter before VV?

The letters before VV that appear in LEVELLEVEL are

EE

and

L.L.

If the first letter is EE, the remaining letters are

L,V,E,L.L,V,E,L.

There are 44 letters, with LL repeated twice, so the number of arrangements is

4!2!=12.\frac{4!}{2!}=12.

If the first letter is LL, the remaining letters are

E,V,E,L.E,V,E,L.

There are 44 letters, with EE repeated twice, so the number of arrangements is

4!2!=12.\frac{4!}{2!}=12.

So the total number of arrangements beginning with a letter before VV is

24.24.

Sum of All Arrangements of Digits

Some problems ask for the sum of all numbers formed by arranging given digits.

The key idea is positional contribution.

If all digits are distinct and none is zero, then each digit appears equally often in each position.

Example 7

Find the sum of all three digit numbers formed using the digits 1,2,31,2,3, with each digit used exactly once.

There are

3!=63!=6

numbers.

Each digit appears in each position

2!2!

times.

The sum of the digits is

1+2+3=6.1+2+3=6.

So the contribution to the hundreds place is

6(2!)100.6(2!)100.

The contribution to the tens place is

6(2!)10.6(2!)10.

The contribution to the units place is

6(2!)1.6(2!)1.

So the total sum is

6(2!)(100+10+1).6(2!)(100+10+1).

Since

2!=2,2!=2,

the sum is

6(2)(111)=1332.6(2)(111)=1332.

Arrangements With Leading Zero

When arranging digits, leading zero is a common issue.

A string like

01230123

is not a four digit number.

So if zero is included, handle the first digit separately or subtract invalid arrangements.

Example 8

How many four digit numbers can be formed using the digits 0,1,2,30,1,2,3, each exactly once?

There are 4!4! arrangements of the four digits.

But those beginning with 00 are not four digit numbers.

If 00 is first, the remaining digits can be arranged in

3!3!

ways.

So the number of valid four digit numbers is

4!−3!=24−6=18.4!-3!=24-6=18.

Key Contest Insight

Arrangement problems are about positions.

Rank problems count how many arrangements come before a given arrangement.

Sum problems count how much each digit contributes to each position.

When the objects are repeated, always divide by the factorials of repeated counts.

Strategy Tips

  • For distinct objects, start with n!n!
  • For repeated objects, divide by the factorials of repeated counts
  • For 'together' restrictions, use blocks
  • For 'not together' restrictions, consider complement counting
  • For rank, count arrangements before the target one position by position
  • For sums of arrangements, use positional contribution
  • Be careful with leading zero

Common Pitfalls

  • Forgetting to divide for repeated letters
  • Treating repeated letters as distinct
  • Forgetting to add 11 when computing rank
  • Counting rank in the wrong alphabetical order
  • Allowing leading zero in a number
  • Trying to list all arrangements instead of using positional contribution

Practice Problems

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