JEE Mains Maths · Permutations and Combinations

Distributions and Integer Solutions

Sharing objects among people or boxes: identical objects (stars and bars, integer solutions with bounds) and distinct objects (onto maps, groups, partitions).

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Why this matters

Seventeen PYQs, and 2024 alone has six. The first question is always whether the objects are identical and whether the boxes are distinct; that decides the method. Eleven share identical objects and six share distinct ones. Two ideas cover the page.

Concept 1 of 2: Identical objects: stars and bars

Sharing

n

identical objects among

k

people is the number of solutions of

x_1+\dots+x_k=n

in non-negative integers:

\binom{n+k-1}{k-1}

. A lower bound is removed by substitution; an upper bound is handled by subtracting the cases that break it. When the boxes are identical too, only the partition of

n

matters, so list the partitions.

Definition

$x_i\ge0$ : $\binom{n+k-1}{k-1}$ ; $x_i\ge1$ : $\binom{n-1}{k-1}$ .
$x_i\ge a_i$ : put $y_i=x_i-a_i$ .
Upper bound $x_i\le b$ : subtract solutions with $x_i\ge b+1$ .
A coefficient such as $a+b+2c=N$ : fix $c$ and add.
Identical into identical boxes: partitions of $n$ .

Stars and bars

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

Worked example

In how many ways can 12 identical sweets be shared among 3 children, each getting at least 2?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q151Moderate

Example 1 · Permutations and Combinations · Distributions and Integer Solutions

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

Distinct values are a separate count

x,y,z

distinct' is not built into stars and bars. Count all solutions, then subtract those with two or three equal values, using inclusion–exclusion.

Drill 10 more on identical objects: stars and bars

Concept 2 of 2: Distinct objects: onto maps and groups

Each of

n

distinct objects chooses one of

k

distinct boxes:

k^n

ways. If no box may be empty, subtract by inclusion–exclusion:

\sum_{j}(-1)^j\binom kj(k-j)^n

. Splitting into groups of given sizes is

\frac{n!}{a!\,b!\cdots}

, divided by the factorial of the number of equal-sized groups if the groups are unlabelled.

Definition

Onto maps: $k^n-\binom k1(k-1)^n+\binom k2(k-2)^n-\cdots$ .
Groups of sizes $a,b,c$ (labelled): $\frac{n!}{a!\,b!\,c!}$ .
Unlabelled equal groups: divide by the factorial of how many are equal.
Into $k$ identical boxes: sum of Stirling numbers $S(n,1)+\dots+S(n,k)$ .

Onto maps

\sum_{j=0}^{k}(-1)^j\binom kj(k-j)^n

Worked example

In how many ways can 4 different balls go into 2 different boxes with neither empty?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q59Moderate

Example 2 · Permutations and Combinations · Distributions and Integer Solutions

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is

Labelled or unlabelled

Cars of different makes are labelled, so which car gets 2 people matters. Unnamed groups of equal size are not: divide by the ways to permute those groups.

Drill 5 more on distinct objects: onto maps and groups

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 (2)

Identical objects: stars and bars
Stars and bars
$x_1+\cdots+x_k=n,\ x_i\ge0:\quad\binom{n+k-1}{k-1}$
Distinct objects: onto maps and groups
Onto maps
$\sum_{j=0}^{k}(-1)^j\binom kj(k-j)^n$

Watch out for (2)

Distinct values are a separate count
→ Identical objects: stars and bars
Labelled or unlabelled
→ Distinct objects: onto maps and groups

Test yourself on Permutations and Combinations

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