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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2019 · Mathematics · Permutation and Combination

If P(n, r)=2520 and C(n, r)=21 , then what is the value of C(n+1,...

If P(n,r)=2520P(n, r)=2520 and C(n,r)=21\mathrm{C}(n, r)=21, then what is the value of C(n+1,r+1)\mathrm{C}(n+1, r+1) ?

Options

  1. A7
  2. B14
  3. C28
  4. D56

Correct answer

C. 28

Step-by-step solution

nPr=2520;nCr=21nPr=n!(n−r)!=2520\begin{aligned} & { }^n P_r=2520 ;{ }^n C_r=21 \\ & { }^n P_r=\frac{n!}{(n-r)!}=2520 \end{aligned} ...(1) nCr=−n!(n−r)!.r!=21{ }^n C_r=-\frac{n!}{(n-r)!. r!}=21 ...(2) Divide (1) by (2) ⇒r!=252021=120⇒r!=5!⇒r=5∵nCr=21⇒nC5=21⇒n=7∴C(n+1,r+1)=C(8,6)=8C6=28\begin{aligned} & \Rightarrow r!=\frac{2520}{21}=120 \\ & \Rightarrow r!=5! \\ & \Rightarrow r=5 \\ & \because{ }^n C_r=21 \\ & \Rightarrow{ }^n C_5=21 \\ & \Rightarrow n=7 \\ & \therefore C(n+1, r+1) \\ & =C(8,6) \\ & ={ }^8 C_6=28 \end{aligned}
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