Formation of Numbers: Playing with Digits
Master the art of counting numbers, calculating sums, and applying divisibility rules using permutations and combinations.
Why Number Formation Matters in JEE
Number formation problems test your understanding of permutations and combinations in practical scenarios. These questions appear regularly in JEE and cover three main areas:
- Counting n-digit numbers with various constraints
- Finding sum of all numbers formed by given digits
- Numbers divisible by specific numbers using divisibility rules
🎯 Topic Navigation
1. Counting n-digit Numbers
Learn to count numbers with different constraints on digits and repetition.
Key Formulas
Without repetition, without 0:
$^nP_r$ or $n \times (n-1) \times \cdots \times (n-r+1)$
Without repetition, with 0:
$(n-1) \times {}^{n-1}P_{r-1}$
With repetition, without 0:
$n^r$
With repetition, with 0:
$(n-1) \times n^{r-1}$
Example 1: Basic Counting
How many 3-digit numbers can be formed from digits {1,2,3,4,5} without repetition?
Step 1: We need to fill 3 positions: _ _ _
Step 2: Hundreds place: 5 choices (1,2,3,4,5)
Step 3: Tens place: 4 choices (remaining digits)
Step 4: Units place: 3 choices (remaining digits)
Step 5: Total = $5 \times 4 \times 3 = 60$
Answer: 60 numbers
Example 2: With Zero Constraint
How many 4-digit numbers can be formed from {0,1,2,3,4,5} without repetition?
Step 1: Thousands place cannot be 0 → 5 choices (1,2,3,4,5)
Step 2: Remaining 3 places from remaining 5 digits: ${}^5P_3 = 5 \times 4 \times 3 = 60$
Step 3: Total = $5 \times 60 = 300$
Answer: 300 numbers
💡 Quick Tips
- Always check if 0 is included in the digit set
- For numbers, the first digit cannot be 0
- Use $^nP_r$ when no repetition is allowed
- Use $n^r$ when repetition is allowed
2. Sum of All Numbers Formed
Learn to calculate the sum of all numbers formed by given digits with/without repetition.
Key Formulas
Sum of all n-digit numbers without repetition:
$\text{Sum} = (\text{Sum of digits}) \times (n-1)! \times (111\ldots1 \text{ n times})$
Sum with repetition allowed:
$\text{Sum} = (\text{Sum of digits}) \times n^{n-1} \times (111\ldots1 \text{ n times})$
When 0 is included:
Adjust for the fact that 0 cannot be the first digit
Example 3: Sum Without Repetition
Find the sum of all 3-digit numbers formed using {1,2,3} without repetition.
Step 1: Sum of digits = 1+2+3 = 6
Step 2: Each digit appears in each position equally
Step 3: Number of times each digit appears in each position = $2! = 2$
Step 4: Sum at each position = $6 \times 2 = 12$
Step 5: Total sum = $12 \times 100 + 12 \times 10 + 12 \times 1 = 1332$
Answer: 1332
Example 4: Sum With Zero
Find the sum of all 3-digit numbers formed using {0,1,2} without repetition.
Step 1: Sum of digits = 0+1+2 = 3
Step 2: Hundreds place: digits appear (n-1)! = 2 times each (except 0)
Step 3: Tens/Units place: each digit appears 2 times
Step 4: Sum at hundreds place = (1+2) × 2 = 6
Step 5: Sum at tens/units place = 3 × 2 = 6
Step 6: Total sum = $6 \times 100 + 6 \times 10 + 6 \times 1 = 666$
Answer: 666
💡 Quick Tips
- Use symmetry - each digit appears equally in each position
- When 0 is included, adjust for hundreds place separately
- The multiplier $(111\ldots1)$ comes from place values
- For n-digit numbers, use $(n-1)!$ for permutations without repetition
3. Numbers Divisible by Specific Numbers
Apply divisibility rules with permutations to count numbers divisible by given numbers.
Key Divisibility Rules
Divisible by 2: Last digit even (0,2,4,6,8)
Divisible by 3: Sum of digits divisible by 3
Divisible by 4: Last two digits divisible by 4
Divisible by 5: Last digit 0 or 5
Divisible by 6: Divisible by both 2 and 3
Divisible by 8: Last three digits divisible by 8
Example 5: Divisible by 2
How many 4-digit numbers can be formed from {1,2,3,4,5} that are divisible by 2?
Step 1: For divisibility by 2, last digit must be even
Step 2: Even digits available: {2,4} → 2 choices
Step 3: First digit: cannot be 0, and cannot be the last digit used
Step 4: Thousands place: 4 choices (remaining digits)
Step 5: Remaining two places: ${}^3P_2 = 3 \times 2 = 6$ choices
Step 6: Total = $2 \times 4 \times 6 = 48$
Answer: 48 numbers
Example 6: Divisible by 4
How many 5-digit numbers from {0,1,2,3,4,5} are divisible by 4?
Step 1: For divisibility by 4, last two digits must form a number divisible by 4
Step 2: Possible last two digits: 04, 12, 20, 24, 32, 40, 52
Step 3: Case 1: Last two digits don't contain 0 → 12, 24, 32, 52 (4 cases)
Step 4: For these: First digit has 3 choices, remaining two places: ${}^3P_2 = 6$
Step 5: Case 2: Last two digits contain 0 → 04, 20, 40 (3 cases)
Step 6: For these: First digit has 3 choices, remaining: ${}^3P_2 = 6$
Step 7: Total = $(4 \times 3 \times 6) + (3 \times 3 \times 6) = 72 + 54 = 126$
Answer: 126 numbers
💡 Quick Tips
- For divisibility by $2^n$, focus on last n digits
- For divisibility by 3, check sum of all digits
- Break complex problems into cases
- Watch for special constraints with 0
📋 Quick Reference Formulas
Counting Numbers
- n-digit without 0: ${}^nP_r$ or $n \times (n-1) \times \cdots$
- n-digit with 0: $(n-1) \times {}^{n-1}P_{r-1}$
- With repetition, no 0: $n^r$
- With repetition, with 0: $(n-1) \times n^{r-1}$
Sum of Numbers
- Without repetition: $(sum) \times (n-1)! \times 111\ldots1$
- With repetition: $(sum) \times n^{n-1} \times 111\ldots1$
- With 0: Adjust hundreds place separately
🎯 Practice Problems
Test your understanding with these JEE-level problems:
1. How many 4-digit even numbers can be formed from {0,1,2,3,4,5} without repetition?
2. Find the sum of all 4-digit numbers formed by {1,3,5,7} without repetition.
3. How many numbers between 1000 and 5000 can be formed from {0,1,2,3,4,5} that are divisible by 5?
⚠️ Common Mistakes to Avoid
Conceptual Errors
- Forgetting that 0 cannot be the first digit in a number
- Confusing permutations vs combinations in counting
- Missing special cases in divisibility problems
- Incorrect application of factorial formulas
Calculation Errors
- Miscounting available choices at each step
- Errors in summation formulas for total sum
- Wrong application of divisibility rules
- Overcounting or undercounting cases
Master Number Formation for JEE Success!
These concepts appear in 1-2 questions in every JEE Main paper