QJEEM25Ma1c56: JEE Main 2025 Mathematics - Class 11 JEEM25Ma1c56 Solution: The number of natural numbers, between 212 and 999, such that…

$\begin{array}{|c|c|c|}\hline \mathrm{x} & \mathrm{y} & \mathrm{z} \\\hline\end{array}$

Let $\mathrm{x}=2 \Rightarrow \mathrm{y}+\mathrm{z}=13$

$(4,9),(5,8),(6,7),(7,6),(8,5),(9,4), \rightarrow 6$

Let $x=3 \rightarrow y+z=12$

$(3,9),(4,8), \ldots \ldots . .,(9,3) \rightarrow 7$

Let $x=4 \rightarrow y+z=11$

$(2,9),(3,8), \ldots \ldots \ldots,(9,1) \rightarrow 9$

Let $x=5 \rightarrow y+z=10$

$(1,9),(2,8), \ldots \ldots . .,(9,1) \rightarrow 10$

Let $x=6 \rightarrow y+z=9$

$(0,9),(1,8), \ldots \ldots . .,(9,0) \rightarrow 9$

Let $\mathrm{x}=7 \rightarrow \mathrm{y}+\mathrm{z}=8$

$(0,9),(1,7), \ldots \ldots . .,(8,0) \rightarrow 9$

Let $x=8 \rightarrow y+z=7$

$(0,7),(1,6), \ldots \ldots . .,(7,0) \rightarrow 8$

Let $x=9 \rightarrow y+z=6$

$(0,6),(1,5), \ldots \ldots \ldots,(6,0) \rightarrow 7$

Total $=6=7+8+9+10+9+8+7=64$