Number Formation with Constraints - Intermediate Level: tricky scenarios handling Number Formation with Constraints INTERMEDIATE

This expert challenge 📈 worksheet focuses on Number Formation with Constraints - 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 number formation with constraints, number formation with constraints tricks, and number formation with constraints 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 Number Formation with Constraints
Worksheet 5 of 10 (44% complete)

Question 1

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 2

How many 5-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 5-digit number).

Given:
- Number length: 5 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 4-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 5th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (4 times)
= 9 × 10^4
= 9 × 10000
= 90000

Alternative Verification:
- Smallest 5-digit number: 10000 = 10000
- Largest 5-digit number: 99999 = 99999
- Total count: 99999 - 10000 + 1 = 90000

Related Problems:
1. No repetition: 9 × P(9,4) = 9 × 9!/5!
2. Odd numbers only: 9 × 10^3 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^3 × 1
- If last digit 2,4,6,8: 8 × 10^3 × 4
- Total: 9 × 10^3 + 8 × 4 × 10^3

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 3

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 4

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 5

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 6

How many 5-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 5-digit number).

Given:
- Number length: 5 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 4-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 5th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (4 times)
= 9 × 10^4
= 9 × 10000
= 90000

Alternative Verification:
- Smallest 5-digit number: 10000 = 10000
- Largest 5-digit number: 99999 = 99999
- Total count: 99999 - 10000 + 1 = 90000

Related Problems:
1. No repetition: 9 × P(9,4) = 9 × 9!/5!
2. Odd numbers only: 9 × 10^3 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^3 × 1
- If last digit 2,4,6,8: 8 × 10^3 × 4
- Total: 9 × 10^3 + 8 × 4 × 10^3

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 7

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 8

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 9

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 10

How many 5-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 5-digit number).

Given:
- Number length: 5 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 4-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 5th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (4 times)
= 9 × 10^4
= 9 × 10000
= 90000

Alternative Verification:
- Smallest 5-digit number: 10000 = 10000
- Largest 5-digit number: 99999 = 99999
- Total count: 99999 - 10000 + 1 = 90000

Related Problems:
1. No repetition: 9 × P(9,4) = 9 × 9!/5!
2. Odd numbers only: 9 × 10^3 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^3 × 1
- If last digit 2,4,6,8: 8 × 10^3 × 4
- Total: 9 × 10^3 + 8 × 4 × 10^3

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 11

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 12

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 13

How many 5-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 5-digit number).

Given:
- Number length: 5 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 4-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 5th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (4 times)
= 9 × 10^4
= 9 × 10000
= 90000

Alternative Verification:
- Smallest 5-digit number: 10000 = 10000
- Largest 5-digit number: 99999 = 99999
- Total count: 99999 - 10000 + 1 = 90000

Related Problems:
1. No repetition: 9 × P(9,4) = 9 × 9!/5!
2. Odd numbers only: 9 × 10^3 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^3 × 1
- If last digit 2,4,6,8: 8 × 10^3 × 4
- Total: 9 × 10^3 + 8 × 4 × 10^3

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 14

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 15

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 16

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 17

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 18

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 19

How many 6-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 6-digit number).

Given:
- Number length: 6 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 5-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 6th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (5 times)
= 9 × 10^5
= 9 × 100000
= 900000

Alternative Verification:
- Smallest 6-digit number: 100000 = 100000
- Largest 6-digit number: 999999 = 999999
- Total count: 999999 - 100000 + 1 = 900000

Related Problems:
1. No repetition: 9 × P(9,5) = 9 × 9!/4!
2. Odd numbers only: 9 × 10^4 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^4 × 1
- If last digit 2,4,6,8: 8 × 10^4 × 4
- Total: 9 × 10^4 + 8 × 4 × 10^4

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices

Question 20

How many 7-digit numbers can be formed using the digits 0 to 9 (with repetition allowed)?
Step-by-Step Solution:

Concept: Number formation with positional restrictions. The first digit cannot be 0 (otherwise it wouldn't be an 7-digit number).

Given:
- Number length: 7 digits
- Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits)
- Repetition: Allowed
- Constraint: First digit cannot be 0

Position-by-Position Analysis:

First digit (leftmost):
Cannot be 0 (would make it 6-digit number)
Choices: 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 9 choices

Second digit:
Can be any digit including 0
Choices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Count: 10 choices

Third digit through 7th digit:
Each can be any digit including 0
Count: 10 choices each

Apply Multiplication Principle:
Total numbers = 9 × 10 × 10 × ... × 10 (6 times)
= 9 × 10^6
= 9 × 1000000
= 9000000

Alternative Verification:
- Smallest 7-digit number: 1000000 = 1000000
- Largest 7-digit number: 9999999 = 9999999
- Total count: 9999999 - 1000000 + 1 = 9000000

Related Problems:
1. No repetition: 9 × P(9,6) = 9 × 9!/3!
2. Odd numbers only: 9 × 10^5 × 5 (last digit: 1,3,5,7,9)
3. Even numbers only:
- If last digit 0: 9 × 10^5 × 1
- If last digit 2,4,6,8: 8 × 10^5 × 4
- Total: 9 × 10^5 + 8 × 4 × 10^5

Key Principle: When forming numbers:
- First digit has special restriction (can't be 0)
- Handle positional constraints carefully
- Use multiplication principle for independent choices
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