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Permutations with Identical Objects - Intermediate Level: tricky scenarios handling Permutations with Identical Objects INTERMEDIATE

This expert challenge 📈 worksheet focuses on Permutations with Identical Objects - a key topic in Permutation Combination. You'll solve 20 intermediate-level problems (Worksheet 5 of 10). The primary focus is on tricky scenarios handling. Master how to solve permutations with identical objects, permutations with identical objects tricks, and permutations with identical objects shortcut methods through systematic practice.

📝 Worksheet 5 of 10 • 20 questions • ⏱️ Estimated time: 20 minutes • 🎯 Intermediate level

What you'll learn in this worksheet:
Your progress through Permutations with Identical Objects
Worksheet 5 of 10 (44% complete)

Question 1

A word has 11 letters: 2 as, 4 bs, 2 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: 11
- Distribution: Type 1: 2, Type 2: 4, Type 3: 2, Type 4: 3

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

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

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

You have 9 books: 3 copies of book type A, 4 copies of book type B, 2 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: 4, Type 3: 2

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

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

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 3

Consider 11 marbles: 4 of color 1, 3 of color 2, 3 of color 3, 1 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: 11
- Distribution: Type 1: 4, Type 2: 3, Type 3: 3, Type 4: 1

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

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

Final Calculation:
= 46200

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 4

A word has 13 letters: 3 as, 4 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: 13
- Distribution: Type 1: 3, Type 2: 4, Type 3: 3, Type 4: 3

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

Step 2 - Account for identical objects:
13! = 6227020800 / 3! = 6 / 4! = 24 / 3! = 6 / 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 5

You have 13 books: 2 copies of book type A, 3 copies of book type B, 4 copies of book type C, 3 copies of book type D, 1 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: 13
- Distribution: Type 1: 2, Type 2: 3, Type 3: 4, Type 4: 3, Type 5: 1

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

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

Final Calculation:
= 3603600

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 6

A word has 18 letters: 4 as, 4 bs, 4 cs, 4 ds, 2 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: 18
- Distribution: Type 1: 4, Type 2: 4, Type 3: 4, Type 4: 4, Type 5: 2

Step 1 - Total arrangements if all were distinct:
18! = 6402373705728000

Step 2 - Account for identical objects:
18! = 6402373705728000 / 4! = 24 / 4! = 24 / 4! = 24 / 4! = 24 / 2! = 2

Final Calculation:
= 9648639000

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 18! = 6402373705728000.

Question 7

You have 7 books: 2 copies of book type A, 3 copies of book type B, 2 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: 7
- Distribution: Type 1: 2, Type 2: 3, Type 3: 2

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

Step 2 - Account for identical objects:
7! = 5040 / 2! = 2 / 3! = 6 / 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 8

You have 8 books: 3 copies of book type A, 2 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: 8
- Distribution: Type 1: 3, Type 2: 2, Type 3: 3

Step 1 - Total arrangements if all were distinct:
8! = 40320

Step 2 - Account for identical objects:
8! = 40320 / 3! = 6 / 2! = 2 / 3! = 6

Final Calculation:
= 560

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 8! = 40320.

Question 9

A word has 12 letters: 2 as, 3 bs, 2 cs, 4 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: 12
- Distribution: Type 1: 2, Type 2: 3, Type 3: 2, Type 4: 4, Type 5: 1

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

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

Final Calculation:
= 831600

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 10

Consider 7 marbles: 2 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: 7
- Distribution: Type 1: 2, Type 2: 2, Type 3: 3

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

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

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 11

A word has 12 letters: 2 as, 2 bs, 4 cs, 3 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: 12
- Distribution: Type 1: 2, Type 2: 2, Type 3: 4, Type 4: 3, Type 5: 1

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

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

Final Calculation:
= 831600

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 12

You have 6 books: 3 copies of book type A, 2 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: 6
- Distribution: Type 1: 3, Type 2: 2, Type 3: 1

Step 1 - Total arrangements if all were distinct:
6! = 720

Step 2 - Account for identical objects:
6! = 720 / 3! = 6 / 2! = 2

Final Calculation:
= 60

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 6! = 720.

Question 13

Consider 10 marbles: 3 of color 1, 2 of color 2, 3 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: 3, Type 2: 2, Type 3: 3, Type 4: 2

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

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

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 14

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 15

Consider 16 marbles: 2 of color 1, 3 of color 2, 4 of color 3, 4 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: 16
- Distribution: Type 1: 2, Type 2: 3, Type 3: 4, Type 4: 4, Type 5: 3

Step 1 - Total arrangements if all were distinct:
16! = 20922789888000

Step 2 - Account for identical objects:
16! = 20922789888000 / 2! = 2 / 3! = 6 / 4! = 24 / 4! = 24 / 3! = 6

Final Calculation:
= 504504000

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 16! = 20922789888000.

Question 16

You have 8 books: 3 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: 8
- Distribution: Type 1: 3, Type 2: 4, Type 3: 1

Step 1 - Total arrangements if all were distinct:
8! = 40320

Step 2 - Account for identical objects:
8! = 40320 / 3! = 6 / 4! = 24

Final Calculation:
= 280

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 8! = 40320.

Question 17

A word has 15 letters: 3 as, 4 bs, 3 cs, 4 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: 15
- Distribution: Type 1: 3, Type 2: 4, 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 / 3! = 6 / 4! = 24 / 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 18

Consider 14 marbles: 3 of color 1, 3 of color 2, 4 of color 3, 3 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: 14
- Distribution: Type 1: 3, Type 2: 3, Type 3: 4, Type 4: 3, Type 5: 1

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

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

Final Calculation:
= 16816800

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 19

A word has 10 letters: 2 as, 4 bs, 2 cs, 2 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: 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 20

Consider 11 marbles: 4 of color 1, 4 of color 2, 2 of color 3, 1 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: 11
- Distribution: Type 1: 4, Type 2: 4, Type 3: 2, Type 4: 1

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

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

Final Calculation:
= 34650

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.
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