Permutation with Identical Objects

Understanding Permutation with Identical Objects

Permutation with identical objects is an arrangement of objects where there are several objects that are identical or the same. When there are identical objects, the number of different arrangements will decrease because exchanging identical objects does not produce new arrangements.

Imagine arranging letters from the word "MAMA". Although there are $4$ letters, we cannot distinguish between the first $M$ and the second $M$, or the first $A$ and the second $A$. As a result, arrangements that look different but use the same letters in different positions are considered identical.

Formula for Identical Object Permutation

For permutation of $n$ objects where there are identical objects, the formula used is:

Explanation:

• $n$ = 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

How to identify identical objects: Count how many times each object appears in the entire arrangement, not just looking at different objects.

Application to Words and Letters

Example Word KALIMANTAN

Let's calculate how many letter arrangements can be made from the word "KALIMANTAN".

Systematic identification steps: Write letters one by one: K-A-L-I-M-A-N-T-A-N

Total letters: $10$

Letter identification: $K$ appears $1 \text{ time}$, $A$ appears $3 \text{ times}$ (positions $2, 6, 9$), $L$ appears $1 \text{ time}$, $I$ appears $1 \text{ time}$, $M$ appears $1 \text{ time}$, $N$ appears $2 \text{ times}$ (positions $7, 10$), and $T$ appears $1 \text{ time}$

Calculation:

Simplify the fraction by canceling common factors:

Calculate step by step:

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

Example Word PALAPA

For the word "PALAPA" with $6$ letters: Write letters one by one: 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: $3 \times 5 \times 4 = 60 \text{ arrangements}$

Systematic Calculation Steps

To solve permutation problems with identical objects, follow these steps:

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

Word BANANA

Let's apply these steps to find arrangements of 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: $3 \times 5 \times 4 = 60$

Therefore, 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

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

2. A flower shop has $8$ roses where $3$ are red, $3$ are white, and $2$ are yellow. How many ways can these flowers be arranged in a row?

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. Total $8$ flowers with red $3$, white $3$, and yellow $2$

   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: $9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3$
   • $= 9 \times 4 \times 7 \times 6 \times 5 \times 4 \times 3 = 90{,}720 \text{ arrangements}$