<p>Find the number of integral solutions of \(x_1 + x_2 + x_3 = 24\) subjected to the condition that \(1 \leq x_1 \leq 5\), \(12 \leq x_2 \leq 18\) and \(-1 \leq x_3\).</p>
Step-by-Step Solution
Key Concept: Transform the constrained problem into an unconstrained one by substituting new variables that shift each original variable into a non-negative range, then apply stars and bars.
<p><strong>Step 1: Change of Variables</strong></p><p>Given: $x_1 + x_2 + x_3 = 24$ with $1 \leq x_1 \leq 5$, $12 \leq x_2 \leq 18$, $-1 \leq x_3$</p><p>Let $y_1 = x_1 - 1$, so $0 \leq y_1 \leq 4$</p><p>Let $y_2 = x_2 - 12$, so $0 \leq y_2 \leq 6$</p><p>Let $y_3 = x_3 + 1$, so $y_3 \geq 0$</p><p><strong>Step 2: Substitute into the Equation</strong></p><p>$(y_1 + 1) + (y_2 + 12) + (y_3 - 1) = 24$</p><p>$y_1 + y_2 + y_3 + 12 = 24$</p><p>$y_1 + y_2 + y_3 = 12$ where $0 \leq y_1 \leq 4$, $0 \leq y_2 \leq 6$, $y_3 \geq 0$</p><p><strong>Step 3: Apply Inclusion-Exclusion</strong></p><p>Without upper bound constraints: $\binom{12+2}{2} = \binom{14}{2} = 91$</p><p>Subtract cases where $y_1 \geq 5$: Let $y_1' = y_1 - 5$. Then $y_1' + y_2 + y_3 = 7$. Count: $\binom{9}{2} = 36$</p><p>Subtract cases where $y_2 \geq 7$: Let $y_2' = y_2 - 7$. Then $y_1 + y_2' + y_3 = 5$. Count: $\binom{7}{2} = 21$</p><p>Add back cases where both $y_1 \geq 5$ AND $y_2 \geq 7$: Then $y_1' + y_2' + y_3 = 0$. Count: $\binom{2}{2} = 1$</p><p><strong>Step 4: Final Count</strong></p><p>$91 - 36 - 21 + 1 = 35$</p><p>∴ <strong>Answer: 35</strong></p>
Correct Answer: 35