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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2014 · Mathematics · Permutation and Combination

Paragraph: Given that C(n, r): C(n, r+1)=1: 2 and C(n, r+1): C(n,...

Paragraph: Given that C(n,r):C(n,r+1)=1:2C(n, r): C(n, r+1)=1: 2 and C(n,r+1):C(n,r+2)=2:3C(n, r+1): C(n, r+2)=2: 3.
Question: What is rr equal to?

Options

  1. A2
  2. B3
  3. C4
  4. D5

Correct answer

C. 4

Step-by-step solution

nCrnCr+1=12(given)⇒n!r!(n−r)!n!(r+1)!(n−r−1)!=12⇒(r+1)!(n−r−1)!r!(n−r)!=12⇒(r+1)⋅r!(n−r−1)!r!(n−r)(n−r−1)!=12⇒n−r=2(r+1)⇒n−r=2r+2⇒n=3r+2..(i) Again, nCr+1nCr+2=23\begin{aligned} & \frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r+1}}=\frac{1}{2} (given)\\ & \Rightarrow \quad \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{1}{2} \\ & \Rightarrow \quad \frac{(r+1)!(n-r-1)!}{r!(n-r)!}=\frac{1}{2} \\ & \Rightarrow \frac{(r+1) \cdot r!(n-r-1)!}{r!(n-r)(n-r-1)!}=\frac{1}{2} \\ & \Rightarrow \quad n-r=2(r+1) \\ & \Rightarrow \quad n-r=2 r+2 \\ & \Rightarrow \quad n=3 r+2 ..(i)\\ & \text{ Again, } \\ & \frac{{ }^n C_{r+1}}{{ }^n C_{r+2}}=\frac{2}{3} \end{aligned}⇒n!(r+1)!(n−r−1)!n!(r+2)!(n−r−2)!=23⇒(r+2)!(n−r−2)!(r+1)!(n−r−1)!=23⇒(r+2)(r+1)!(n−r−2)!(r+1)!(n−r−1)(n−r−2)!=232(n−r−1)=3(r+2)⇒2n−2r−2=3r+6⇒2n=5r+8...(ii)\begin{aligned} & \Rightarrow \quad \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{(r+2)!(n-r-2)!}}=\frac{2}{3} \\ & \Rightarrow \quad \frac{(r+2)!(n-r-2)!}{(r+1)!(n-r-1)!}=\frac{2}{3} \\ & \Rightarrow \frac{(r+2)(r+1)!(n-r-2)!}{(r+1)!(n-r-1)(n-r-2)!}=\frac{2}{3} \\ & 2(n-r-1)=3(r+2) \\ & \Rightarrow \quad 2 n-2 r-2=3 r+6 \\ & \Rightarrow \quad 2 n=5 r+8...(ii) \end{aligned}Multiplyequation(i)by2Multiply equation (i) by 22 n=6 r+4...(iii)Fromequations(ii)and(iii)...(iii) From equations (ii) and (iii)6r+4=5r+8⇒r=4\begin{aligned} & & 6 r+4 & =5 r+8 \\ \Rightarrow & & r & =4 \end{aligned}PuttingthevalueofPutting the value ofrinequation(ii)in equation (ii)2n=5(4)+82n=20+82n=28n=14.\begin{aligned} 2 n & =5(4)+8 \\ 2 n & =20+8 \\ 2 n & =28 \\ n & =14 . \end{aligned}ValueofValue ofr=4$.
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