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Question Id : 11505 | Context :NIMCET 2023

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
🎥 Video solution / Text Solution of this question is given below:

Candidate Passing Criteria

Total Papers: 9

Condition for Success: Passes > Fails

So, candidate is unsuccessful when: Passes ≤ 4

Calculate ways:

\text{Ways} = \sum_{x=0}^{4} \binom{9}{x} = \binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4} = 1 + 9 + 36 + 84 + 126 = \boxed{256}

✅ Final Answer: 256 ways


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