Permutation and CombinationHard
Question
A student has to answer 10 out of 13 questions in an examination. The number of ways in which he can answer if he must answer atleast 3 of the first five questions is:
Options
A.276
B.267
C.13C10 - 5C3
D.5C3 . 8C7 + 5C4 . 8C6 + 8C5
Solution
Total number of required possibilities
5C3 . 8C7 + 5C4 . 8C6 + 5C5 . 8C5 . 5C5
= 5C3 . 8C7 + 5C4 . 8C6 + 8C6
= 13C10 - 5C3 = 276
5C3 . 8C7 + 5C4 . 8C6 + 5C5 . 8C5 . 5C5
= 5C3 . 8C7 + 5C4 . 8C6 + 8C6
= 13C10 - 5C3 = 276
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