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NDA Maths · Permutation & Combination

Forming Numbers from Digits

Counting how many numbers can be built from given digits under constraints — number of digits, a leading-zero rule, divisibility, or the sum of all such numbers.

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Why this matters

Digit problems are permutations dressed in number rules. The two perennial gotchas are the leading-zero exclusion and divisibility tests; the 'sum of all numbers' shortcut saves real time.

Concept 1 of 3

Counting numbers with digit constraints

Intuition

Building an nn-digit number is filling nn ordered slots from the available digits. The one special rule: the leading digit can't be 0, so fill it first from the non-zero digits.

Definition

Fill positions left to right. Distinct digits: first slot has (non-zero choices), each later slot one fewer. Leading-zero rule: if 0 is available, the first digit has one fewer option; equivalently total arrangements minus those starting with 0. Repetition allowed multiplies the per-slot choices.

Worked example

How many 3-digit numbers have all distinct digits from {1,2,3,4}\{1,2,3,4\}?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Permutation & CombinationEASY
How many 4-digit numbers are there having all digits as odd?

[Q3 · Sep · 2024]

Drill 10 more on counting numbers with digit constraints

Concept 2 of 3

Divisibility constraints

Intuition

Divisibility fixes specific digits. By 2/5/10 → the units digit; by 4 → the last two digits; by 3/9 → the digit sum. Often the digit sum is fixed (e.g. 1+2+3+4+5=151+2+3+4+5=15), which forces divisibility by 3 for every arrangement.

Definition

  • ÷2: units even. ÷5: units 0 or 5. ÷10: units 0.
  • ÷4: last two digits form a multiple of 4. ÷8: last three.
  • ÷3 / ÷9: digit sum divisible by 3 / 9 (independent of order — so a fixed digit set is all-or-nothing).
  • ÷6: divisible by 2 and 3 together.

Worked example

How many 3-digit numbers from 1,2,3,4,51,2,3,4,5 (no repeat) are divisible by 5?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Permutation & CombinationMODERATE
How many 4-digit numbers that are divisible by 4 can be formed using the digits 1, 2, 3 and 4 (repetition of digits is not allowed)?

[Q18 · Sep · 2025]

Drill 6 more on divisibility constraints

Concept 3 of 3

Sum of all numbers formed

Intuition

By symmetry, each digit lands in each position the same number of times. So the sum of all numbers formed is (sum of digits) × (times each appears per place) × (place-value repunit).

Definition

Using nn distinct digits to form all nn-digit numbers: each digit appears in each place (n−1)!(n-1)! times. Sum =(n−1)!×(sum of digits)×111…1⏟n=(n-1)!\times(\text{sum of digits})\times\underbrace{111\ldots1}_{n}. Adjust the repeat count and place-value string for rr-digit selections.

Sum of all numbers formed from n distinct digits

Sum=(n−1)!×(sum of digits)×111…1⏟n ones\text{Sum}=(n-1)!\times(\text{sum of digits})\times\underbrace{111\ldots1}_{n\text{ ones}}

Worked example

Find the sum of all 3-digit numbers formed using 3,4,53,4,5 without repetition.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Permutation & CombinationMODERATE
What is the sum of all three-digit numbers that can be formed using all the digits 3, 4 and 5, when repetition of digits is not\textbf{\text{not}} allowed?

[Q48 · Sep · 2018]

Drill 1 more on sum of all numbers formed

Summary — formulas & gotchas at a glance

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Formulas (1)

  • Sum of all numbers formed

    Sum of all numbers formed from n distinct digits

    Sum=(n−1)!×(sum of digits)×111…1⏟n ones\text{Sum}=(n-1)!\times(\text{sum of digits})\times\underbrace{111\ldots1}_{n\text{ ones}}

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