JEE Mains Maths · Permutations and Combinations

Forming Numbers from Digits

Counting numbers built from a given set of digits: with a range or position condition, with a divisibility condition, or with a condition on the sum or product of the digits.

All Permutations and Combinations notes JEE Mains Maths strategy

Why this matters

Thirty-seven PYQs, the largest page in the chapter, and twenty-seven of them numerical answer. Sixteen fill positions under a range or no-repetition rule, fourteen need a divisibility test, and seven fix the sum or product of the digits. Three ideas cover the page.

Concept 1 of 3: Filling positions: ranges, leading digits and no repetition

Fill the most restricted position first — the leading digit (never 0, bounded by the range), the last digit (parity) — then the rest. A range like 'between 5000 and 9000' fixes the leading digit's choices. To add all the numbers formed, find how often each digit sits in each place: by symmetry, every place carries the same digit total.

Definition

Restricted positions first, then the rest in order.
No repetition: choices drop by one at each step.
Sum of all numbers from $n$ distinct digits (all used): $(n-1)!\,(\text{digit sum})\,(11\dots1)$ .
Three-digit numbers with exactly one digit repeated twice: all, minus all digits equal, minus all digits distinct.

Sum of all arrangements

\sum=\frac{(\text{number of arrangements})\times(\text{digit sum})}{n}\times\underbrace{11\cdots1}_{n}

Worked example

How many 4-digit numbers greater than 6000 can be formed from 2, 4, 6, 8, 9 without repetition?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q169Moderate

Example 1 · Permutations and Combinations · Forming Numbers from Digits

The number of integers, greater than 7000 that can be formed, using the digits

3,5,6,7,8

without repetition, is

Zero in the leading place

When 0 is available, the leading digit has one fewer choice. If the last digit is also restricted (say even, and 0 is even), split into cases 'last digit 0' and 'last digit not 0'.

Drill 15 more on filling positions: ranges, leading digits and no repetition

Concept 2 of 3: Divisibility conditions

Turn each divisibility rule into a condition on digits: by 5, last digit 0 or 5; by 4, the last two digits; by 3, the digit sum; by 11, the alternating sum. For 'divisible by 3' with repetition allowed, group the digits by remainder mod 3 and count the combinations of remainders that add to a multiple of 3. Combined divisors (6, 15, 55) need both tests.

Definition

By 3 or 9: digit sum. By 4: last two digits. By 5: last digit 0 or 5.
By 11: (sum of odd places) $-$ (sum of even places) is a multiple of 11.
Six-digit palindromes $\overline{abccba}$ are all multiples of 11.
With repetition, if the allowed digits are spread evenly over remainders mod 3, exactly a third of the free choices work.

Digit-sum test

3\mid\overline{d_1d_2\cdots d_n}\ \text{exactly when}\ 3\mid d_1+d_2+\cdots+d_n

Worked example

How many 4-digit numbers divisible by 4 can be formed from 1, 6, 8, 9 without repetition?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q84Moderate

Example 2 · Permutations and Combinations · Forming Numbers from Digits

The number of 3-digit numbers, formed using the digits

2,3,4,5

and 7 , when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to

Count the complement when it is smaller

'Not divisible by 3' is usually easier as total minus divisible. Listing the digit triples whose sum is a multiple of 3 is short; listing the others is long.

Drill 13 more on divisibility conditions

Concept 3 of 3: Fixed digit sum or product

A fixed digit sum is an equation

d_1+\dots+d_n=S

in bounded integers: substitute to remove the lower bounds, count with stars and bars, then subtract the cases where a digit exceeds 9. A fixed product is a factorisation: list the multisets of digits with that product, then count arrangements of each.

Definition

$a+b+c=S$ , $a\ge1$ , $b,c\ge0$ : put $a'=a-1$ , then $\binom{S-1+2}{2}$ before bounds.
Subtract each case with a digit $\ge10$ (put $d'=d-10$ ).
Digits from $\{1,2,3\}$ with sum $S$ : solve for the counts of each digit, then $\frac{n!}{a!\,b!\,c!}$ .
Product $k$ : list digit multisets whose product is $k$ .

Stars and bars

x_1+\cdots+x_k=S,\ x_i\ge0:\ \binom{S+k-1}{k-1}

Worked example

How many 3-digit numbers have digit sum 5?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q174Moderate

Example 3 · Permutations and Combinations · Forming Numbers from Digits

The number of integers, between 100 and 1000 having the sum of their digits equals to 14 , is

Digits stop at 9

Stars and bars counts solutions with any size of digit. For a sum above 9, remove the solutions where some digit is 10 or more before answering.

Drill 6 more on fixed digit sum or product

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Filling positions: ranges, leading digits and no repetition
Sum of all arrangements
$\sum=\frac{(\text{number of arrangements})\times(\text{digit sum})}{n}\times\underbrace{11\cdots1}_{n}$
Divisibility conditions
Digit-sum test
$3\mid\overline{d_1d_2\cdots d_n}\ \text{exactly when}\ 3\mid d_1+d_2+\cdots+d_n$
Fixed digit sum or product
Stars and bars
$x_1+\cdots+x_k=S,\ x_i\ge0:\ \binom{S+k-1}{k-1}$

Watch out for (3)

Zero in the leading place
→ Filling positions: ranges, leading digits and no repetition
Count the complement when it is smaller
→ Divisibility conditions
Digits stop at 9
→ Fixed digit sum or product

Test yourself on Permutations and Combinations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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