Question

(1) The sum of all the values of $r$ satisfying ${}^{39}{C}_{3r-1}{-}^{39}{C}_{r^2}{=}^{39}{C}_{r^2-1}{-}^{39}{C}_{3r}\text{is}{\alpha }_1.$

(2) If ${}^{2n+3}{C}_{2n}{-}^{2n+2}{C}_{2n-1}=15.(2n+1)$ then the value of $n$ is ${\alpha }_2.$

(3) If ${}^{56}{P}_{r+6}{:}^{54}{P}_{r+3}=30800:1$ then value of $r$ is ${\alpha }_3.$

(4) ${}^{n+2}{C}_8{:}^{n-2}{P}_4=57:16$ then the value of $n$ is ${\alpha }_4.$

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(I\right)& \text{The value of}{\alpha }_1\text{is}& \left(P\right)& 41\\ \left(II\right)& \text{The value of}{\alpha }_2\text{is}& \left(Q\right)& 8\\ \left(III\right)& \text{The value of}{\alpha }_3\text{is}& \left(R\right)& 14\\ \left(IV\right)& \text{The value of}{\alpha }_4\text{is}& \left(S\right)& 19\end{array}$

$\text{Which of the following is only INCORRECT Combination?}$

• A. $\left(I\right)\to \left(Q\right)$
• B. $\left(II\right)\to \left(P\right)$
• C. $\left(III\right)\to \left(P\right)$
• D. $\left(IV\right)\to \left(S\right)$

Answer 1

The correct option is B $\left(II\right)\to \left(P\right)$

(I) Given that,
${}^{39}{C}_{3r-1}{-}^{39}{C}_{r^2}{=}^{39}{C}_{r^2-1}{-}^{39}{C}_{3r}{\Rightarrow }^{39}{C}_{3r-1}{+}^{39}{C}_{3r}{=}^{39}{C}_{r^2-1}{+}^{39}{C}_{r^2}{\Rightarrow }^{40}{C}_{3r}{=}^{40}{C}_{r^2}\therefore \text{Either}{r}^2=3r\Rightarrow r=3\text{or}{r}^2=40-\left(3r\right)\Rightarrow r^2+3r-40=0\Rightarrow r=5\text{Sum of possible values of}=3+5=8={\alpha }_1$

(II) Given that,
${}^{2n+3}{C}_{2n}{-}^{2n+2}{C}_{2n-1}=15.(2n+1){\Rightarrow }^{2n+2}{C}_{2n}{+}^{2n+2}{C}_{2n-1}{-}^{2n+2}{C}_{2n-1}=15.(2n+1){\Rightarrow }^{2n+2}{C}_{2n}=15.(2n+1)\Rightarrow \frac{(2n+2)!}{\left(2n\right)!\times 2!}=15.(2n+1)\Rightarrow (2n+2)(2n+1)=30.(2n+1)\Rightarrow n+1=15\Rightarrow n=14={\alpha }_2$

(III) Given that,
${}^{56}{P}_{r+6}{:}^{54}{P}_{r+3}=30800:1\Rightarrow \frac{56!}{(50-r)!}\times \frac{(51-r)!}{54!}=\frac{30800}{1}\Rightarrow 56\times 55\times (51-r)=30800\Rightarrow 51-r=10\Rightarrow r=41={\alpha }_3$

(IV) Given that,
${}^{n+2}{C}_8{:}^{n-2}{P}_4=57:16\Rightarrow \frac{(n+2)!}{8!\times (n-6)!}\times \frac{(n-6)!}{(n-2)!}=\frac{57}{16}\Rightarrow (n+2)(n+1)n(n-1)=143640\Rightarrow (n+2)(n+1)n(n-1)=21\times 20\times 19\times 18\Rightarrow n=19={\alpha }_4$