Question

how to solve this question????​

Answer 1

$\textsf{Here \ m\ \&\ n \ are positive integers.}$

$\displaystyle \begin{aligned}& \ ^{m+n}\!P_2=56\\ \\ \Longrightarrow\ \ &\frac{(m+n)!}{(m+n-2)!}=56\\ \\ \Longrightarrow\ \ &\frac{1 \times 2 \times 3 \times ...... (m+n-2)(m+n-1)(m+n)}{1 \times 2 \times 3 \times ...... (m+n-2)}\\ \\ \Longrightarrow\ \ &(m+n-1)(m+n)=56\end{aligned}$

$\textsf{On taking \ m+n=x , }\\ \\ \\ \begin{aligned}\Longrightarrow\ \ &(x-1)x=56\\ \\ \\ \\ \Longrightarrow\ \ &x^2-x=56\\ \\ \Longrightarrow\ \ &x^2-x-56=0\\ \\ \Longrightarrow\ \ &x^2-8x+7x-56=0\\ \\ \Longrightarrow\ \ &x(x-8)+7(x-8)=0\\ \\ \Longrightarrow\ \ &(x+7)(x-8)=0\\ \\ \\ \therefore\ \ &x=-7\ \ \ ; \ \ \ x=8\\ \\ \Longrightarrow\ \ &m+n=-7\ \ \ ; \ \ \ m+n=8\end{aligned}\\ \\ \\ \textsf{But as m and n are positive integers, }\ m+n=8\ \ \ \longrightarrow\ \ \ (1)$

$\displaystyle \begin{aligned}&\ ^{m-n}\!P_3=24\\ \\ \Longrightarrow\ \ &\frac{(m-n)!}{(m-n-3)!}=24\\ \\ \Longrightarrow\ \ &\frac{1 \times 2 \times 3 \times .......(m-n-3)(m-n-2)(m-n-1)(m-n)}{1 \times 2 \times 3 \times ...... (m-n-3)}=24\\ \\ \Longrightarrow\ \ &(m-n-2)(m-n-1)(m-n)=24\end{aligned}$

$\textsf{On taking \ m-n=x , }\\ \\ \begin{aligned}\Longrightarrow\ \ &(x-2)(x-1)x=24\\ \\ \Longrightarrow\ \ &x^3-3x^2+2x=24\\ \\ \Longrightarrow\ \ &x^3-3x^2+2x-24=0\\ \\ \Longrightarrow\ \ &x^3-4x^2+x^2-4x+6x-24=0\\ \\ \Longrightarrow\ \ &x^2(x-4)+x(x-4)+6(x-4)=0\\ \\ \Longrightarrow\ \ &(x-4)(x^2+x+6)=0\end{aligned}$

$\textsf{As \ x^2+x+6 \ has no integer roots, }\\ \\ \begin{aligned}&x=4\\ \\ \Longrightarrow\ \ &m-n=4\ \ \ \longrightarrow\ \ \ (2)\end{aligned}$

$\begin{aligned}&(1)+(2)\\ \\ \Longrightarrow\ \ &(m+n)+(m-n)=8+4\\ \\ \Longrightarrow\ \ &2m=12\\ \\ \Longrightarrow\ \ &m=6\\ \\ \\ &(1)-(2)\\ \\ \Longrightarrow\ \ &(m+n)-(m-n)=8-4\\ \\ \Longrightarrow\ \ &2n=4\\ \\ \Longrightarrow\ \ &n=2\end{aligned}$

$\displaystyle \ ^mP_3=\ ^6P_3=\frac{6!}{3!}=6 \times 5 \times 4=120\\ \\ \\ \ ^nP_2=\ ^2P_2=\frac{2!}{0!}=2$

$\displaystyle \begin{aligned}&\textsf{Now, }\\ \\ &\frac{\ ^mP_3}{\ ^nP_2}\\ \\ \Longrightarrow\ \ &\frac{120}{2}\\ \\ \Longrightarrow\ \ &\large \bold{60}\end{aligned}$