Permutations & Combinations
Non-negative integral solutions
Grade 11
Question:
<p>Let <em>n</em> and <em>k</em> be positive integers such that \( n \geq \dfrac{k(k+1)}{2} \). Find the number of solutions \((x_1, x_2, \ldots, x_k)\), \(x_1 \geq 1,\, x_2 \geq 2,\, \ldots,\, x_k \geq k\), all integers satisfying the condition \(x_1 + x_2 + x_3 + \cdots + x_k = n\).</p>
Step-by-Step Solution
Key Concept: Transform the constrained problem into an unconstrained one by substituting $y_i = x_i - i$ for each variable, converting lower bounds $x_i \geq i$ into $y_i \geq 0$, then apply stars and bars to count non-negative integer solutions.
<p><strong>Step 1: Identify the constraint structure.</strong> We have $x_1 + x_2 + \cdots + x_k = n$ with $x_i \geq i$ for $i = 1, 2, \ldots, k$.</p><p><strong>Step 2: Apply substitution.</strong> Let $y_i = x_i - i$ for each $i$. Then $x_i = y_i + i$ and $y_i \geq 0$ (integers).</p><p><strong>Step 3: Rewrite the equation in terms of $y_i$.</strong> Substituting into the sum:
$$\sum_{i=1}^{k}(y_i + i) = n$$
$$\sum_{i=1}^{k}y_i + \sum_{i=1}^{k}i = n$$
$$\sum_{i=1}^{k}y_i = n - \frac{k(k+1)}{2}$$</p><p><strong>Step 4: Apply stars and bars.</strong> The number of non-negative integer solutions to $y_1 + y_2 + \cdots + y_k = m$ is $\binom{m+k-1}{k-1}$, where $m = n - \frac{k(k+1)}{2}$.</p><p><strong>Step 5: Express the final answer.</strong>
$$\binom{m+k-1}{k-1} = \frac{(m+k-1)!}{(k-1)! \cdot m!}$$
where $m = n - \frac{k(k+1)}{2}$.</p><p>∴ Answer: $\dfrac{(m+k-1)!}{(k-1)!\, m!}$ where $m = n - \dfrac{k(k+1)}{2}$</p>
Correct Answer: \(\dfrac{(m+k-1)!}{(k-1)!\, m!}\) where \(m = n - \dfrac{k(k+1)}{2}\)