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Question: Answered & Verified by Expert
A student is asked to answer 10 out of 13 questions in an examination such that he must answer atleast four questions from the first five questions. Then the total number of possible choices available to him is
MathematicsPermutation CombinationTS EAMCETTS EAMCET 2023 (14 May Shift 1)
Options:
  • A $186$
  • B $176$
  • C $286$
  • D $196$
Solution:
1352 Upvotes Verified Answer
The correct answer is: $196$
$\left({ }^5 \mathrm{C}_4 \times{ }^8 \mathrm{C}_6\right)+\left({ }^5 \mathrm{C}_5 \times{ }^8 \mathrm{C}_5\right)$
$=(5 \times 28)+(1 \times 56)=196$

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