Number of integral solutions to the equation $x+y+z=21$, where $x \ge 1,y\ge3,z\ge4$, is equal to ____________.
Answer (integer)
105
Solution
$\begin{aligned} & x+y+z=21 \\\\ & \because \quad x \geq 1, y \geq 3, y \geq 4 \\\\ & \therefore \quad x_1+y_1+z_1=13 \\\\ & \text { Number of solutions }={ }^{13+3-1} C_{3-1} \\\\ & ={ }^{15} C_2=\frac{15 \times 14}{2}=7 \times 15 \\\\ & =105\end{aligned}$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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