Let n and k be positive such that n ≥ k(k+1)/2. The number of solutions (x₁, x₂, ..., xₖ), x₁ ≥ 1, x

Q: Why can't we just use the standard Stars and Bars formula directly on $n$?

The standard Stars and Bars formula $\binom{n+k-1}{k-1}$ only applies when variables are non-negative ($x_i \ge 0$). Here, your variables have different lower bounds ($x_1 \ge 1, x_2 \ge 2, \dots$). You must shift the variables to make them non-negative first.

Q: What if $n < \frac{k(k+1)}{2}$? Does the formula still work?

If $n < \frac{k(k+1)}{2}$, it means the total number of items is less than the minimum required to satisfy the constraints. In this case, the number of solutions is zero. The formula will naturally yield a negative value or a combination that is undefined, which correctly reflects that no solutions exist.

Q: Is there a way to solve this using generating functions?

Yes! You could find the coefficient of $x^n$ in the expansion of $(x^1 + x^2 + \dots)(x^2 + x^3 + \dots)\dots(x^k + x^{k+1} + \dots)$. However, this is significantly more complex than the variable-shifting method we used. The Stars and Bars approach is the most elegant and efficient path for this type of problem.

Question

Let $n$ and $k$ be positive such that $n \ge \frac{k(k+1)}{2}$. The number of solutions $(x_1, x_2, \dots, x_k)$, $x_1 \ge 1, x_2 \ge 2, \dots, x_k \ge k$, all integers, satisfying $x_1 + x_2 + \dots + x_k = n$, is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with solving the equation $x_1 + x_2 + \dots + x_k = n$, subject to the constraints $x_1 \ge 1, x_2 \ge 2, \dots, x_k \ge k$.

At first glance, this appears to be a standard counting problem, but the varying lower bounds act as hidden traps. To solve this, we must systematically account for these constraints.

### The Philosophy of Pre-allocation

Imagine you have a pool of $n$ identical coins to distribute among $k$ people, where each person has a specific minimum demand. Person 1 requires at least $1$ coin, Person 2 requires at least $2$, and so on, up to Person $k$ who requires at least $k$ coins.

If we distribute coins randomly, we will likely fail to meet these demands. The secret is to pay the "tax" first by pre-allocating the minimum required amounts to each person.

The total number of coins we must hand out immediately is the sum of the first $k$ natural numbers:

$$1 + 2 + \dots + k = \frac{k(k+1)}{2}$$

This represents our mandatory investment.

### The Transformation

Now, let us translate this physical intuition into the language of algebra. We have a total pool of $n$ coins. After paying the tax of $\frac{k(k+1)}{2}$, the remaining pool of coins, which we denote as $N$, is:

$$N = n - \frac{k(k+1)}{2}$$

These $N$ coins are now free to be distributed without any further restrictions. To formalize this, we define new variables $y_i = x_i - i$.

Since $x_i \ge i$, it follows that $y_i \ge 0$. We have successfully transformed a problem with complex constraints into a problem with the simplest possible constraint: non-negativity.

### The Power of Stars and Bars

Our original equation $x_1 + x_2 + \dots + x_k = n$ transforms into:

$$(y_1 + 1) + (y_2 + 2) + \dots + (y_k + k) = n$$

Rearranging this, we obtain:

$$y_1 + y_2 + \dots + y_k = n - (1 + 2 + \dots + k)$$

This simplifies to:

$$y_1 + y_2 + \dots + y_k = n - \frac{k(k+1)}{2}$$

This is the classic **Stars and Bars** scenario. The theorem states that the number of non-negative integer solutions to the sum of $k$ variables equaling $N$ is given by the binomial coefficient $\binom{N + k - 1}{k - 1}$.

### Final Calculation

We now substitute our value of $N$ back into the formula. The total number of solutions is:

$$\binom{n - \frac{k(k+1)}{2} + k - 1}{k - 1}$$

Take a moment to appreciate the elegance of this result. We started with a problem that seemed restrictive and difficult, and through the simple act of shifting our variables, we reduced it to a standard combinatorial form.

This is the essence of **JEE-level problem solving**: not just calculating, but transforming the problem into a form that reveals its own solution.

Visualized Solution

The Original Equation

Visualizing the Total Pool

Fulfilling Minimum Demands

Calculating the Remaining Items

Mathematical Substitution

Forming the New Equation

The Stars and Bars Theorem

The Final Solution

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Philosophy of Pre-allocation

The Transformation

The Power of Stars and Bars

Final Calculation

Similar Questions

Number of integral solutions to the equation $x + y + z = 21$ , where $x \geq 1, y \geq 3, z \geq 4$ , is equal to ____.

The number of triplets $(x, y, z)$ where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$ , is

Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is

The total number of functions, $f : {1, 2, 3, 4} \to {1, 2, 3, 4, 5, 6}$ such that $f (1) + f (2) = f (3)$ , is equal to:

The number of bijective functions $f : {1, 3, 5, 7, \dots, 99} \to {2, 4, 6, 8, \dots, 100}$ , such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq \dots \geq f (99)$ , is

Let $A$ be a $3 \times 3$ matrix having entries from the set ${- 1, 0, 1}$ . The number of all such matrices $A$ having sum of all the entries equal to 5, is

The total number of positive integral solutions $(x, y, z)$ such that $x y z = 24$ is

Let $S = {1, 2, 3, 5, 7, 10, 11}$ . The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ____.

Let $S = {1, 2, 3, 4}$ . The total number of unordered pairs of disjoint subsets of $S$ is equal to

If $^{n} C_{r}$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression $^{n} C_{r + 1} +^{n} C_{r - 1} + 2 \times^{n} C_{r}$ equals