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The number of solutions of the equation $x+y+z=10$ where $x, y$ and $z$ are positive integers
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Verified Answer
The correct answer is:
36
Given equation, is $x+y+z=10$
where, $x, y$ and $z$ are positive integers.
$\therefore$ Required number of solutions $={ }^{(10-1)} \mathrm{C}_{(3-1)}$
$$
={ }^{9} C_{2}=\frac{9 \times 8}{2}=36
$$
where, $x, y$ and $z$ are positive integers.
$\therefore$ Required number of solutions $={ }^{(10-1)} \mathrm{C}_{(3-1)}$
$$
={ }^{9} C_{2}=\frac{9 \times 8}{2}=36
$$
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