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6.Permutation and Combination
easy
In an examination there are three multiple choice questions and each question has $4 $ choices. Number of ways in which a student can fail to get all answers correct, is
A
$11$
B
$12$
C
$27$
D
$63$
Solution
(d) Each question can be answered in $4$ ways and all questions can be answered correctly in only one way, so required number of ways $ = {4^3} – 1 = 63$.
Standard 11
Mathematics