Step-by-step Solution:

The candidate passes in some papers and fails in others.
To be successful, the number of passes must be strictly greater than the number of fails.
To be unsuccessful, the number of fails must be greater than or equal to the number of passes.
Let:
\( p \) be the number of papers the candidate passes.
\( f \) be the number of papers the candidate fails.
Since there are 9 papers: \[ p + f = 9 \] For the candidate to be unsuccessful: \[ f \geq p \] This implies: \[ f \geq \frac{9}{2} = 4.5 \] Since \( f \) is an integer, the possible values are: \[ f \in \{5,6,7,8,9\} \] Each paper has two possibilities (pass or fail), so the total number of ways to attempt the 9 papers is: \[ 2^9 = 512 \] Since exactly half of these cases correspond to failing in more than half of the subjects (symmetry in binomial choices), we get: \[ \frac{512}{2} = 256 \] \Final Answer: \[ {256} \]