Question

In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is:

• A. 255
• B. 256
• C. 128
• D. \( 9 \times 8! \)

Hint:

To calculate the number of ways a candidate can be unsuccessful in an exam, consider the different cases based on the number of failures and apply combinations to count the number of possible outcomes.

Answer 1

The Correct Option is B

In the examination, there are 9 papers, and the candidate must pass in more papers than the number of papers he fails in. 

- There are two possibilities for each paper: either he passes or he fails. 

- The total number of possible outcomes (ways of passing or failing) for all 9 papers is \( 2^9 = 512 \). 

To be unsuccessful, the candidate must fail in at least 5 papers (because he needs to pass in more papers than he fails). Thus, we need to find the number of outcomes in which he fails in 5 or more papers. 

- If he fails in 5 papers, there are \( \binom{9}{5} \) ways to choose which 5 papers he fails. 

- Similarly, for failing in 6, 7, 8, or 9 papers, the number of ways are \( \binom{9}{6}, \binom{9}{7}, \binom{9}{8}, \binom{9}{9} \). 

The total number of ways he can be unsuccessful is the sum of all these outcomes: \[ \binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9} \] This sum is equal to \( 256 \), which is the total number of ways he can be unsuccessful. 

Thus, the correct answer is \( \boxed{256} \).