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.

All Permutation & Combination notes NDA Maths strategy

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

n

-digit number is filling

n

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\}

No zero present, so just $^4P_3=4\times3\times2$ .

Answer:

24

Practice this conceptself-check · 4 quick reps

Try it yourself

How many numbers greater than 1000 can be formed using

0,1,2,3

without repetition?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Which digit cannot be 0?
2.
3-digit distinct numbers from $\{1,2,3,4\}$ ?
3.
Handle leading zero by?
4.
Repetition allowed: slots are?

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=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

1,2,3,4,5

(no repeat) are divisible by 5?

Divisible by 5 ⇒ units digit is 5 (only choice here): 1 way.
First two slots from the remaining 4 digits: $4\times3=12$ .

Answer:

12

Practice this conceptself-check · 4 quick reps

Try it yourself

How many 5-digit primes can be formed using all of

1,2,3,4,5

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Divisible by 4 depends on?
2.
Divisible by 3 depends on?
3.
Divisible by 10 needs units digit?
4.
5-digit numbers from 1–5: divisible by 3?

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

Worked example

Find the sum of all 3-digit numbers formed using

3,4,5

without repetition.

Each digit appears in each place $(3-1)!=2$ times; digit sum $=12$ .
Sum $=2\times12\times111=2664$ .

Answer:

2664

Practice this conceptself-check · 4 quick reps

Try it yourself

Each digit of

\{1,2,3\}

appears how many times in the units place across all 3-digit numbers (no repeat)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Each digit appears per place how many times (n distinct)?
2.
Place-value string for 3-digit sums?
3.
Sum of all 3-digit numbers from 3,4,5?
4.
Sum formula factors?

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

\textbf{\text{not}}

allowed?

[Q48 · Sep · 2018]

Drill 1 more on sum of all numbers formed

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Permutation & CombinationEASY

Three-digit numbers are formed from the digits 1, 2 and 3 in such a way that the digits are not repeated. What is the sum of such three-digit numbers?

[Q7 · Apr · 2017]

Example 2Permutation & CombinationMODERATE

How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?

[Q50 · Apr · 2021]

Example 3Permutation & CombinationHARD

What is the sum of all four-digit numbers formed by using all digits

0,1,4,5

without repetition of digits?

[Q3 · Apr · 2024]

Example 4Permutation & CombinationMODERATE

How many numbers greater than 1000 can be formed using the digits 0, 1, 2 and 3 (repetition of digits is not allowed)?

[Q44 · Apr · 2026]

Example 5Permutation & CombinationHARD

Four digit numbers are formed by using the digits

1,2,3,5

without repetition of digits. How many of them are divisible by 4?

[Q17 · Apr · 2024]

Drill every past-year question on this subtopic

20 questions from the bank — paginated, with cart and Word-export support.

Drill the 20 questions

Drill every past-year question on this subtopic

Related notes