Advanced Integral Solutions with Constraints | JEE Combinatorics

Beyond Basic Integral Solutions

While $x_1 + x_2 + \cdots + x_r = n$ with $x_i \geq 0$ is straightforward, JEE Advanced introduces complex constraints like:

Lower bounds: $x_i \geq a_i$ (minimum values)
Upper bounds: $x_i \leq b_i$ (maximum values)
Mixed constraints: Different bounds for different variables
Non-homogeneous bounds: $a \leq x_i \leq b$ with $a \neq 0$

🎯 JEE Advanced Focus

These problems test your understanding of inclusion-exclusion principle, generating functions, and complementary counting. Mastery can earn you 8-12 marks in every JEE Advanced paper.

Technique 1 Medium

Lower Bounds Only: The Substitution Method

Problem Type

Find number of integral solutions to $x_1 + x_2 + \cdots + x_r = n$ with $x_i \geq a_i$

Solution Approach

Step 1: Substitute $y_i = x_i - a_i$ for each variable

Step 2: This transforms the equation to:

$y_1 + y_2 + \cdots + y_r = n - (a_1 + a_2 + \cdots + a_r)$

Step 3: With $y_i \geq 0$

Step 4: Number of solutions = $\binom{n - \sum a_i + r - 1}{r - 1}$

Example

Find number of solutions to $x_1 + x_2 + x_3 = 15$ with $x_1 \geq 2, x_2 \geq 3, x_3 \geq 4$

Substitute: $y_1 = x_1 - 2, y_2 = x_2 - 3, y_3 = x_3 - 4$

New equation: $y_1 + y_2 + y_3 = 15 - (2+3+4) = 6$

Solutions: $\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28$

Technique 2 Hard

Upper Bounds Only: Complementary Counting

Problem Type

Find number of integral solutions to $x_1 + x_2 + \cdots + x_r = n$ with $x_i \leq b_i$

Solution Approach

Step 1: Start with total solutions without upper bounds

Step 2: Subtract solutions where at least one $x_i > b_i$

Step 3: Use inclusion-exclusion principle

Step 4: For each $x_i > b_i$, substitute $y_i = x_i - (b_i + 1)$

Example

Find number of solutions to $x_1 + x_2 + x_3 = 10$ with $x_i \geq 0$ and $x_1 \leq 3, x_2 \leq 4, x_3 \leq 5$

Total solutions: $\binom{10 + 3 - 1}{3 - 1} = \binom{12}{2} = 66$

Case A1: $x_1 \geq 4$, let $y_1 = x_1 - 4$

Equation: $y_1 + x_2 + x_3 = 6$ → $\binom{8}{2} = 28$

Case A2: $x_2 \geq 5$, let $y_2 = x_2 - 5$

Equation: $x_1 + y_2 + x_3 = 5$ → $\binom{7}{2} = 21$

Case A3: $x_3 \geq 6$, let $y_3 = x_3 - 6$

Equation: $x_1 + x_2 + y_3 = 4$ → $\binom{6}{2} = 15$

Intersections: Subtract double counts, add triple counts

Final: $66 - (28 + 21 + 15) + (\text{intersections}) = 6$

Technique 3 Hard

Both Lower and Upper Bounds

Problem Type

Find number of integral solutions to $x_1 + x_2 + \cdots + x_r = n$ with $a_i \leq x_i \leq b_i$

Solution Approach

Step 1: First handle lower bounds using substitution

Step 2: Then apply upper bounds using inclusion-exclusion

Step 3: Combine both techniques systematically

Example

Find number of solutions to $x_1 + x_2 + x_3 = 20$ with $2 \leq x_1 \leq 6, 3 \leq x_2 \leq 7, 4 \leq x_3 \leq 8$

Handle lower bounds: $y_1 = x_1 - 2, y_2 = x_2 - 3, y_3 = x_3 - 4$

New equation: $y_1 + y_2 + y_3 = 20 - 9 = 11$ with $y_i \geq 0$

Upper bounds become: $y_1 \leq 4, y_2 \leq 4, y_3 \leq 4$

Now apply inclusion-exclusion for upper bounds...

Technique 4 Advanced

Inclusion-Exclusion Principle Mastery

The General Formula

For $x_1 + x_2 + \cdots + x_r = n$ with $0 \leq x_i \leq b_i$:

$N = \sum_{k=0}^r (-1)^k \sum_{1 \leq i_1 < \cdots < i_k \leq r} \binom{n - \sum_{j=1}^k (b_{i_j} + 1) + r - 1}{r - 1}$

Step-by-Step Application

Step 1: Start with total solutions: $\binom{n + r - 1}{r - 1}$

Step 2: Subtract solutions where each $x_i > b_i$

Step 3: Add back solutions where two variables exceed bounds

Step 4: Continue alternating signs

Step 5: Stop when the binomial coefficient becomes invalid (negative top)

Example

Find number of solutions to $x_1 + x_2 + x_3 + x_4 = 18$ with $0 \leq x_i \leq 6$

Total: $\binom{18 + 4 - 1}{3} = \binom{21}{3} = 1330$

Single exceed: $x_i \geq 7$, 4 cases × $\binom{18-7+4-1}{3} = 4 × \binom{14}{3} = 4 × 364 = 1456$

Double exceed: $\binom{4}{2} × \binom{18-14+4-1}{3} = 6 × \binom{7}{3} = 6 × 35 = 210$

Triple exceed: $\binom{4}{3} × \binom{18-21+4-1}{3} = 4 × \binom{0}{3} = 0$

Final: $1330 - 1456 + 210 - 0 = 84$

📋 Quick Reference Table

Constraint Type	Method	Formula/Approach	When to Use
$x_i \geq a_i$	Substitution	$\binom{n - \sum a_i + r - 1}{r - 1}$	Only lower bounds
$x_i \leq b_i$	Inclusion-Exclusion	Alternating sum formula	Only upper bounds
$a_i \leq x_i \leq b_i$	Combination	Substitution + Inclusion-Exclusion	Both bounds present
Large $n$, small $b_i$	Generating Functions	Coefficient of $x^n$ in $\prod (1+x+\cdots+x^{b_i})$	Complex bounds

Technique 5 Advanced

Generating Functions Approach

The Power of GFs

For $x_1 + x_2 + \cdots + x_r = n$ with $a_i \leq x_i \leq b_i$:

$G(x) = \prod_{i=1}^r (x^{a_i} + x^{a_i+1} + \cdots + x^{b_i})$

Number of solutions = Coefficient of $x^n$ in $G(x)$

Simplification Tricks

Geometric series: $x^a + x^{a+1} + \cdots + x^b = x^a \cdot \frac{1 - x^{b-a+1}}{1 - x}$

Product form: $G(x) = \frac{\prod_{i=1}^r (x^{a_i} - x^{b_i+1})}{(1 - x)^r}$

Binomial expansion: Use $(1-x)^{-r} = \sum_{k=0}^\infty \binom{r+k-1}{k} x^k$

Example

Find number of solutions to $x_1 + x_2 = 8$ with $1 \leq x_1 \leq 5, 2 \leq x_2 \leq 6$

Generating function: $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 + x^5 + x^6)$

Simplify: $x(1+x+x^2+x^3+x^4) \cdot x^2(1+x+x^2+x^3+x^4) = x^3 (1+x+x^2+x^3+x^4)^2$

Coefficient of $x^8$: Coefficient of $x^5$ in $(1+x+x^2+x^3+x^4)^2$

Ways to get $x^5$: $(x,x^4), (x^2,x^3), (x^3,x^2), (x^4,x)$ → 4 solutions

Technique 6 Medium

Complementary Counting for Bounded Variables

When to Use This

When upper bounds are small and it's easier to count violations than valid solutions

Strategy

Step 1: Count total solutions without upper bounds

Step 2: Count solutions where at least one bound is violated

Step 3: Subtract using inclusion-exclusion

Step 4: This is often easier than direct counting

⚠️ Common Pitfall

Students forget that when a variable exceeds its upper bound, we need to count all such cases, not just exactly at the bound.

🎯 JEE Advanced Practice Problems

1. Find number of solutions to $x_1 + x_2 + x_3 + x_4 = 15$ with $0 \leq x_i \leq 6$

Hint: Use inclusion-exclusion with 4 variables

2. Find number of solutions to $x_1 + x_2 + x_3 = 12$ with $x_1 \geq 2, x_2 \geq 3, x_3 \geq 4$ and $x_1 \leq 5, x_2 \leq 6, x_3 \leq 7$

Hint: Handle lower bounds first, then upper bounds

3. Using generating functions, find number of solutions to $x_1 + x_2 + x_3 = 10$ with $1 \leq x_i \leq 4$

Hint: Find coefficient of $x^{10}$ in $(x+x^2+x^3+x^4)^3$

🚀 JEE Exam Strategy

Time Management

Simple bounds: 2-3 minutes max
Medium complexity: 4-5 minutes
Advanced bounds: 6-8 minutes
If stuck after 3 minutes, move on and return

Method Selection

Only lower bounds: Always use substitution
Only upper bounds: Inclusion-exclusion
Small n, simple bounds: Direct enumeration
Complex bounds: Generating functions

Master These 6 Techniques for JEE Success!

Integral solutions with constraints appear in every JEE Advanced paper. Complete mastery can earn you 8-12 guaranteed marks.

Integral Solutions Revisited: Advanced Constraints

Beyond Basic Integral Solutions

🎯 JEE Advanced Focus

🔍 Technique Navigation

Lower Bounds Only: The Substitution Method

Problem Type

Solution Approach

Example

Upper Bounds Only: Complementary Counting

Problem Type

Solution Approach

Example

Both Lower and Upper Bounds

Problem Type

Solution Approach

Example

Inclusion-Exclusion Principle Mastery

The General Formula

Step-by-Step Application

Example

📋 Quick Reference Table

Generating Functions Approach

The Power of GFs

Simplification Tricks

Example

Complementary Counting for Bounded Variables

When to Use This

Strategy

⚠️ Common Pitfall

🎯 JEE Advanced Practice Problems

🚀 JEE Exam Strategy

Time Management

Method Selection

Master These 6 Techniques for JEE Success!