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(C) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the marks assigned to the $4$ questions respectively. We are given the equation: $x_{1} + x_{2} + x_{3} + x_{4

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$10$

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(C) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the marks assigned to the $4$ questions respectively. We are given the equation: $x_{1} + x_{2} + x_{3} + x_{4} = 10$, where $x_{i} \geq 2$ for all $i \in \{1, 2, 3, 4\}$. Let $y_{i} = x_{i} - 2$. Since $x_{i} \geq 2$, we have $y_{i} \geq 0$. Substituting $x_{i} = y_{i} + 2$ into the equation: $(y_{1} + 2) + (y_{2} + 2) + (y_{3} + 2) + (y_{4} + 2) = 10$ $y_{1} + y_{2} + y_{3} + y_{4} + 8 = 10$ $y_{1} + y_{2} + y_{3} + y_{4} = 2$. The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$, where $n = 2$ and $r = 4$. Number of ways $= \binom{2+4-1}{4-1} = \binom{5}{3} = \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = 10$.

What is the number of ways in which an examiner can assign $10$ marks to $4$ questions, giving not less than $2$ marks to any question?

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