The number of ways in which an examiner can a | vedclass

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(A) Let $n_i$ be the marks assigned to the $i^{th}$ question,where $i = 1, 2, \dots, 8$.We are given $n_1 + n_2 + \dots + n_8 = 30$ with $n_i \ge 2$.L

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$^{21}C_{7}$

Vedclass · Answer

(A) Let $n_i$ be the marks assigned to the $i^{th}$ question,where $i = 1, 2, \dots, 8$.We are given $n_1 + n_2 + \dots + n_8 = 30$ with $n_i \ge 2$.Let $x_i = n_i - 2$,so $x_i \ge 0$.Substituting this into the sum: $(x_1 + 2) + (x_2 + 2) + \dots + (x_8 + 2) = 30$.$x_1 + x_2 + \dots + x_8 + 16 = 30$,which simplifies to $x_1 + x_2 + \dots + x_8 = 14$.The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 14$ and $r = 8$.Number of ways = $\binom{14+8-1}{8-1} = \binom{21}{7} = ^{21}C_{7}$.

The number of ways in which an examiner can assign $30$ marks to $8$ questions,awarding not less than $2$ marks to any question is

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