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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Permutation and Combination

BITSAT 2024 · Paper 2 · Mathematics · Permutation and Combination

At an election a voter may vote for any number of candidates not...

At an election a voter may vote for any number of candidates not exceeding the number to be elected. If 4 candidates are to be elected out of the 12 contested in the election and voter votes for at least one candidate, then the number of way of selections

Options

  1. A793793
  2. B298298
  3. C781781
  4. D15851585

Correct answer

A. 793793

Step-by-step solution

Total contested candidates =12=12 4 candidates are to be elected and voter votes for at least one candidate then total number of ways of selections. =12C1+12C2+12C3+12C4=12+12×112×1+12×11×103×2×1+12×11×10×94×3×2×1=12+66+220+495=793={ }^{12} \mathrm{C}_1+{ }^{12} \mathrm{C}_2+{ }^{12} \mathrm{C}_3+{ }^{12} \mathrm{C}_4=12+\frac{12 \times 11}{2 \times 1}+\frac{12 \times 11 \times 10}{3 \times 2 \times 1}+\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}=12+66+220+495=793
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