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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

If ^ 22 P_ r+1 : ^ 20 P_ r+2 =11: 52 , then r is equal to

If 22Pr+1:20Pr+2=11:52{ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52, then rr is equal to

Options

  1. A33
  2. B55
  3. C77
  4. D99

Correct answer

C. 77

Step-by-step solution

{ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52 or (21−r)(20−r)(19−r)=52×2×21⇒(21−r)(20−r)(19−r)=14×13×12(21-r)(20-r)(19-r)=52 \times 2 \times 21\Rightarrow(21-r)(20-r)(19-r)=14 \times 13 \times 12⇒(21−r)(20−r)(19−r)=(21−7)(20−7)(19−7)\begin{aligned} & \Rightarrow(21-r)(20-r)(19-r)=(21-7)(20-7)(19 \\ & -7)\end{aligned}⇒r=7\Rightarrow r=7
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