Permutations with Identical Objects

Why identical objects change the count

A permutation with identical objects counts arrangements when some objects cannot be distinguished from one another. The count is smaller than an ordinary permutation because exchanging identical objects does not produce a new arrangement.

Consider the letters in "MAMA". There are $4$ letters, but the two copies of $M$ are indistinguishable, as are the two copies of $A$. If we temporarily label the copies, swapping those labels would be counted as a different permutation even though the visible word stays the same.

Formula for Permutations with Identical Objects

For $n$ objects divided into groups of identical objects, the number of distinct arrangements is:

Here, each symbol has a specific meaning:

• $n$ = total number of objects
• $r_1, r_2, r_3, \ldots, r_k$ = number of identical objects in each group
• $k$ = number of groups of identical objects
• The frequencies include every object, so $r_1 + r_2 + \cdots + r_k = n$

Count how often each distinct object appears. Dividing by each $r_i!$ removes the repeated counts caused by exchanging identical copies within that group.

Application to Words and Letters

The same division works for written words, because a repeated letter creates identical copies. The example below counts the letters of KALIMANTAN and divides by the repeated groups.

Example Word KALIMANTAN

Calculate how many letter arrangements can be made from the word "KALIMANTAN".

Write the letters in order: K-A-L-I-M-A-N-T-A-N

Total letters: $10$

The repeated letters are $A$, which appears $3 \text{ times}$, and $N$, which appears $2 \text{ times}$. Every other letter appears once.

Calculation:

Simplify the fraction by canceling common factors:

Finish the calculation:

• Divide $10$ by $2$: $\frac{10}{2} = 5$
• The remaining product is $5 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 302\,400$

Example Word PALAPA

For the word "PALAPA" with $6$ letters: Write the letters in order: P-A-L-A-P-A

Letter identification: $P$ appears $2 \text{ times}$ (positions $1, 5$), $A$ appears $3 \text{ times}$ (positions $2, 4, 6$), and $L$ appears $1 \text{ time}$

Calculate each factorial:

• $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
• $2! = 2 \times 1 = 2$
• $3! = 3 \times 2 \times 1 = 6$

So, the calculation is:

Simplify by dividing $6$ by $2$:

• $\frac{6}{2} = 3$
• So: $3 \times 5 \times 4 = 60 \text{ arrangements}$

Counting Arrangements with Repeated Objects

Use the same procedure for any collection that contains repeated objects:

1. Count total objects: Determine the value of $n$
2. Identify identical objects: Group objects that are identical
3. Count frequency: Determine how many times each object appears
4. Apply formula: Insert into the permutation formula
5. Calculate factorial: Complete the calculation carefully

Arrangements of BANANA

Apply the procedure to the word "BANANA":

1. Count total objects

   Write letters one by one: B-A-N-A-N-A

   Total letters: $n = 6$

2. Identify identical objects

   Group identical letters together:

   • B group: B
   • A group: A, A, A
   • N group: N, N

3. Count frequency

   Count how many times each letter appears:

   • $B$ appears $1 \text{ time}$
   • $A$ appears $3 \text{ times}$
   • $N$ appears $2 \text{ times}$

4. Apply formula

   Use the permutation formula with identical objects:

   $$P = \frac{n!}{r_1! \times r_2! \times r_3!}$$

   $$P = \frac{6!}{1! \times 3! \times 2!}$$

5. Calculate factorial

   Simplify the fraction first:

   $$P = \frac{6!}{1! \times 3! \times 2!}$$

   $$= \frac{6 \times 5 \times 4 \times 3!}{1! \times 3! \times 2!}$$

   $$= \frac{6 \times 5 \times 4}{1 \times 2!}$$

   $$= \frac{6 \times 5 \times 4}{2}$$

   Calculate with simplification:

   • Divide $6$ by $2$: $\frac{6}{2} = 3$
   • So: $3 \times 5 \times 4 = 60$

The word "BANANA" can be arranged in $60$ different ways.

Difference from Regular Permutation

Regular permutation: All objects are different, using formula $n!$

Permutation with identical objects: There are identical objects, using formula:

Comparison example:

Arranging letters A, B, C, D (all different): $4! = 24 \text{ ways}$

Arranging letters $A, A, B, C$ (some identical):

Identical objects reduce the number of arrangements because exchanging identical objects does not produce differences.

Exercises

Each problem arranges a multiset, so repeated letters, colors, or digits must be divided out. Count the total arrangements first, then divide by the factorials of the repeated groups.

1. How many letter arrangements can be made from the word "MATEMATIKA"?

2. A flower shop arranges $8$ roses in a row. Only color matters: $3$ are red, $3$ are white, and $2$ are yellow. How many color arrangements are possible?

3. From the digits $1, 1, 2, 2, 2, 3$, how many $6$-digit numbers can be formed?

4. How many different letter arrangements does the word "INDONESIA" have?

Answer Key

1. The word "MATEMATIKA" has $10$ letters

   Letters one by one: M-A-T-E-M-A-T-I-K-A

   Letter identification: $M$ appears $2 \text{ times}$ (positions $1, 5$), $A$ appears $3 \text{ times}$ (positions $2, 6, 10$), $T$ appears $2 \text{ times}$ (positions $3, 7$), $E$ appears $1 \text{ time}$, $I$ appears $1 \text{ time}$, and $K$ appears $1 \text{ time}$

   Simplify the fraction by canceling common factors:

   $$P = \frac{10!}{2! \times 3! \times 2! \times 1! \times 1! \times 1!}$$

   $$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3!}{2! \times 3! \times 2!}$$

   $$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{2! \times 2!}$$

   $$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{4}$$

   Calculate with simplification:

   • Complete calculation: $10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 604\,800$
   • Divide by $4$: $\frac{604\,800}{4} = 151\,200 \text{ arrangements}$

2. There are $8$ color positions: red appears $3$ times, white appears $3$ times, and yellow appears $2$ times.

   Simplify the fraction by canceling common factors:

   $$P = \frac{8!}{3! \times 3! \times 2!}$$

   $$= \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3!}{3! \times 3! \times 2!}$$

   $$= \frac{8 \times 7 \times 6 \times 5 \times 4}{3! \times 2!}$$

   $$= \frac{8 \times 7 \times 6 \times 5 \times 4}{6 \times 2}$$

   $$= \frac{8 \times 7 \times 6 \times 5 \times 4}{12}$$

   Calculate with simplification:

   • Divide $6$ by $12$: $\frac{6}{12} = \frac{1}{2}$
   • So: $8 \times 7 \times \frac{1}{2} \times 5 \times 4$
   • $= 4 \times 7 \times 5 \times 4 = 560 \text{ ways}$

3. Digits $1, 1, 2, 2, 2, 3$ (total $6$ digits)

   Digit identification: digit $1$ appears $2 \text{ times}$, digit $2$ appears $3 \text{ times}$, and digit $3$ appears $1 \text{ time}$

   Simplify the fraction by canceling common factors:

   $$P = \frac{6!}{2! \times 3! \times 1!}$$

   $$= \frac{6 \times 5 \times 4 \times 3!}{2! \times 3!}$$

   $$= \frac{6 \times 5 \times 4}{2!}$$

   $$= \frac{6 \times 5 \times 4}{2}$$

   Calculate with simplification:

   • Divide $6$ by $2$: $\frac{6}{2} = 3$
   • So: $3 \times 5 \times 4 = 60 \text{ numbers}$

4. The word "INDONESIA" has $9$ letters

   Letters one by one: I-N-D-O-N-E-S-I-A

   Letter identification: $I$ appears $2 \text{ times}$ (positions $1, 8$), $N$ appears $2 \text{ times}$ (positions $2, 5$), $D$ appears $1 \text{ time}$, $O$ appears $1 \text{ time}$, $E$ appears $1 \text{ time}$, $S$ appears $1 \text{ time}$, and $A$ appears $1 \text{ time}$

   Simplify the fraction by canceling common factors:

   $$P = \frac{9!}{2! \times 2! \times 1! \times 1! \times 1! \times 1! \times 1!}$$

   $$= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2! \times 2!}$$

   $$= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2!}$$

   $$= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{2}$$

   Calculate with simplification:

   • Divide by $2$: $\frac{8}{2} = 4$
   • So: $9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3$
   • $= 9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3 = 90\,720 \text{ arrangements}$