NIMCET 2023 Question Paper with solutions

Question : 9 of 120

Marks: +1, -0

In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in which he fails in order to get the success. The number of ways in which he can be fail, is:

255
256
$9 × 8$ !
128

Validate

Solution:

The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers

∴

The number of ways to be unsuccessful

\;={ }^9 {\C}_9+{ }^9 {\C}_8+{ }^9 {\C}_7+{ }^9 {\C}_6+{ }^9 {\C}_5

\;={ }^9 {\C}_0+{ }^9 {\C}_1+{ }^9 {\C}_2+{ }^9 {\C}_3+{ }^9 {\C}_4

\;=\;{1}/{2} ({ }^9 {\C}_0+{ }^9 {\C}_1+. . .+{ }^9 {\C}_9)

\;=\;{1}/{2} × 2{ }^9=256 .

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