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Permutations with Identical Objects: Worksheet 10 - Expert Practice Permutations with Identical Objects EXPERT

Ready to master Permutations with Identical Objects? This accuracy focus 👑 worksheet (10/10) presents 20 expert-level challenges. Focus area: application-based learning. Learn to solve permutations with identical objects reasoning tricks, handle fast permutations with identical objects solving, and perfect permutations with identical objects mastery with our step-by-step solutions.

📝 Worksheet 10 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Expert level

What you'll learn in this worksheet:
Your progress through Permutations with Identical Objects
Worksheet 10 of 10 (100% complete)

Question 1

You have 11 books: 3 copies of book type A, 4 copies of book type B, 2 copies of book type C, 2 copies of book type D. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 11
- Distribution: Type 1: 3, Type 2: 4, Type 3: 2, Type 4: 2

Step 1 - Total arrangements if all were distinct:
11! = 39916800

Step 2 - Account for identical objects:
11! = 39916800 / 3! = 6 / 4! = 24 / 2! = 2 / 2! = 2

Final Calculation:
= 69300

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 11! = 39916800.

Question 2

Consider 12 marbles: 3 of color 1, 3 of color 2, 4 of color 3, 2 of color 4. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 12
- Distribution: Type 1: 3, Type 2: 3, Type 3: 4, Type 4: 2

Step 1 - Total arrangements if all were distinct:
12! = 479001600

Step 2 - Account for identical objects:
12! = 479001600 / 3! = 6 / 3! = 6 / 4! = 24 / 2! = 2

Final Calculation:
= 277200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 12! = 479001600.

Question 3

You have 12 books: 4 copies of book type A, 2 copies of book type B, 3 copies of book type C, 3 copies of book type D. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 12
- Distribution: Type 1: 4, Type 2: 2, Type 3: 3, Type 4: 3

Step 1 - Total arrangements if all were distinct:
12! = 479001600

Step 2 - Account for identical objects:
12! = 479001600 / 4! = 24 / 2! = 2 / 3! = 6 / 3! = 6

Final Calculation:
= 277200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 12! = 479001600.

Question 4

A word has 14 letters: 4 as, 2 bs, 3 cs, 2 ds, 3 es. How many distinct ways can these letters be arranged?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 14
- Distribution: Type 1: 4, Type 2: 2, Type 3: 3, Type 4: 2, Type 5: 3

Step 1 - Total arrangements if all were distinct:
14! = 87178291200

Step 2 - Account for identical objects:
14! = 87178291200 / 4! = 24 / 2! = 2 / 3! = 6 / 2! = 2 / 3! = 6

Final Calculation:
= 25225200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 14! = 87178291200.

Question 5

You have 11 books: 4 copies of book type A, 3 copies of book type B, 2 copies of book type C, 2 copies of book type D. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 11
- Distribution: Type 1: 4, Type 2: 3, Type 3: 2, Type 4: 2

Step 1 - Total arrangements if all were distinct:
11! = 39916800

Step 2 - Account for identical objects:
11! = 39916800 / 4! = 24 / 3! = 6 / 2! = 2 / 2! = 2

Final Calculation:
= 69300

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 11! = 39916800.

Question 6

Consider 10 marbles: 2 of color 1, 4 of color 2, 2 of color 3, 2 of color 4. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 10
- Distribution: Type 1: 2, Type 2: 4, Type 3: 2, Type 4: 2

Step 1 - Total arrangements if all were distinct:
10! = 3628800

Step 2 - Account for identical objects:
10! = 3628800 / 2! = 2 / 4! = 24 / 2! = 2 / 2! = 2

Final Calculation:
= 18900

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 10! = 3628800.

Question 7

Consider 15 marbles: 4 of color 1, 3 of color 2, 3 of color 3, 4 of color 4, 1 of color 5. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 15
- Distribution: Type 1: 4, Type 2: 3, Type 3: 3, Type 4: 4, Type 5: 1

Step 1 - Total arrangements if all were distinct:
15! = 1307674368000

Step 2 - Account for identical objects:
15! = 1307674368000 / 4! = 24 / 3! = 6 / 3! = 6 / 4! = 24

Final Calculation:
= 63063000

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 15! = 1307674368000.

Question 8

A word has 10 letters: 2 as, 2 bs, 3 cs, 3 ds. How many distinct ways can these letters be arranged?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 10
- Distribution: Type 1: 2, Type 2: 2, Type 3: 3, Type 4: 3

Step 1 - Total arrangements if all were distinct:
10! = 3628800

Step 2 - Account for identical objects:
10! = 3628800 / 2! = 2 / 2! = 2 / 3! = 6 / 3! = 6

Final Calculation:
= 25200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 10! = 3628800.

Question 9

You have 9 books: 4 copies of book type A, 4 copies of book type B, 1 copies of book type C. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 9
- Distribution: Type 1: 4, Type 2: 4, Type 3: 1

Step 1 - Total arrangements if all were distinct:
9! = 362880

Step 2 - Account for identical objects:
9! = 362880 / 4! = 24 / 4! = 24

Final Calculation:
= 630

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 9! = 362880.

Question 10

Consider 14 marbles: 2 of color 1, 2 of color 2, 4 of color 3, 3 of color 4, 3 of color 5. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 14
- Distribution: Type 1: 2, Type 2: 2, Type 3: 4, Type 4: 3, Type 5: 3

Step 1 - Total arrangements if all were distinct:
14! = 87178291200

Step 2 - Account for identical objects:
14! = 87178291200 / 2! = 2 / 2! = 2 / 4! = 24 / 3! = 6 / 3! = 6

Final Calculation:
= 25225200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 14! = 87178291200.

Question 11

Consider 5 marbles: 2 of color 1, 2 of color 2, 1 of color 3. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 5
- Distribution: Type 1: 2, Type 2: 2, Type 3: 1

Step 1 - Total arrangements if all were distinct:
5! = 120

Step 2 - Account for identical objects:
5! = 120 / 2! = 2 / 2! = 2

Final Calculation:
= 30

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 5! = 120.

Question 12

You have 9 books: 3 copies of book type A, 3 copies of book type B, 3 copies of book type C. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 9
- Distribution: Type 1: 3, Type 2: 3, Type 3: 3

Step 1 - Total arrangements if all were distinct:
9! = 362880

Step 2 - Account for identical objects:
9! = 362880 / 3! = 6 / 3! = 6 / 3! = 6

Final Calculation:
= 1680

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 9! = 362880.

Question 13

You have 12 books: 2 copies of book type A, 3 copies of book type B, 3 copies of book type C, 2 copies of book type D, 2 copies of book type E. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 12
- Distribution: Type 1: 2, Type 2: 3, Type 3: 3, Type 4: 2, Type 5: 2

Step 1 - Total arrangements if all were distinct:
12! = 479001600

Step 2 - Account for identical objects:
12! = 479001600 / 2! = 2 / 3! = 6 / 3! = 6 / 2! = 2 / 2! = 2

Final Calculation:
= 1663200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 12! = 479001600.

Question 14

Consider 13 marbles: 3 of color 1, 3 of color 2, 4 of color 3, 3 of color 4. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 13
- Distribution: Type 1: 3, Type 2: 3, Type 3: 4, Type 4: 3

Step 1 - Total arrangements if all were distinct:
13! = 6227020800

Step 2 - Account for identical objects:
13! = 6227020800 / 3! = 6 / 3! = 6 / 4! = 24 / 3! = 6

Final Calculation:
= 1201200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 13! = 6227020800.

Question 15

A word has 7 letters: 3 as, 2 bs, 2 cs. How many distinct ways can these letters be arranged?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 7
- Distribution: Type 1: 3, Type 2: 2, Type 3: 2

Step 1 - Total arrangements if all were distinct:
7! = 5040

Step 2 - Account for identical objects:
7! = 5040 / 3! = 6 / 2! = 2 / 2! = 2

Final Calculation:
= 210

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 7! = 5040.

Question 16

A word has 14 letters: 4 as, 4 bs, 3 cs, 2 ds, 1 es. How many distinct ways can these letters be arranged?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 14
- Distribution: Type 1: 4, Type 2: 4, Type 3: 3, Type 4: 2, Type 5: 1

Step 1 - Total arrangements if all were distinct:
14! = 87178291200

Step 2 - Account for identical objects:
14! = 87178291200 / 4! = 24 / 4! = 24 / 3! = 6 / 2! = 2

Final Calculation:
= 12612600

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 14! = 87178291200.

Question 17

You have 12 books: 3 copies of book type A, 2 copies of book type B, 4 copies of book type C, 3 copies of book type D. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 12
- Distribution: Type 1: 3, Type 2: 2, Type 3: 4, Type 4: 3

Step 1 - Total arrangements if all were distinct:
12! = 479001600

Step 2 - Account for identical objects:
12! = 479001600 / 3! = 6 / 2! = 2 / 4! = 24 / 3! = 6

Final Calculation:
= 277200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 12! = 479001600.

Question 18

You have 10 books: 2 copies of book type A, 4 copies of book type B, 3 copies of book type C, 1 copies of book type D. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 10
- Distribution: Type 1: 2, Type 2: 4, Type 3: 3, Type 4: 1

Step 1 - Total arrangements if all were distinct:
10! = 3628800

Step 2 - Account for identical objects:
10! = 3628800 / 2! = 2 / 4! = 24 / 3! = 6

Final Calculation:
= 12600

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 10! = 3628800.

Question 19

Consider 9 marbles: 4 of color 1, 2 of color 2, 3 of color 3. How many distinct ways can these marbles be arranged in a row?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 9
- Distribution: Type 1: 4, Type 2: 2, Type 3: 3

Step 1 - Total arrangements if all were distinct:
9! = 362880

Step 2 - Account for identical objects:
9! = 362880 / 4! = 24 / 2! = 2 / 3! = 6

Final Calculation:
= 1260

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 9! = 362880.

Question 20

You have 12 books: 2 copies of book type A, 3 copies of book type B, 2 copies of book type C, 3 copies of book type D, 2 copies of book type E. How many distinct ways can these books be lined on a shelf?
Step-by-Step Solution:

Concept: Permutations with identical objects. When objects of the same type are indistinguishable, we divide by the factorial of each type's count.

Formula:
$$\frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!}$$
where $n$ is total objects and $n_i$ is the count of type $i$.

Given Data:
- Total objects: 12
- Distribution: Type 1: 2, Type 2: 3, Type 3: 2, Type 4: 3, Type 5: 2

Step 1 - Total arrangements if all were distinct:
12! = 479001600

Step 2 - Account for identical objects:
12! = 479001600 / 2! = 2 / 3! = 6 / 2! = 2 / 3! = 6 / 2! = 2

Final Calculation:
= 1663200

Key Principle: Each group of identical objects overcounts by a factor of (count)!. Division corrects this.

Verification: The result is an integer and less than 12! = 479001600.
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