Question 2 Exercise 6.3

Solutions of Question 2 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find $n$ and $r$ if ${ }^n P_r=840$ and ${ }^n C_r=35$.

We are given: $$\begin{align} &^n P_r=\dfrac{n !}{(n-r) !}=840 ....(i)\\ &^n C_r=\dfrac{n !}{(n-r) ! r !}=35....(ii)\end{align}$$ Dividing Eq.(i) by Eq.(ii) $$\begin{align}\dfrac{n !}{(n-r) !} \cdot \dfrac{(n-r) ! r !}{n !}&=\dfrac{840}{35}\\ r!&=24\\ \text{or}\quad r &=4\end{align}$$ Putting $r=4$ in Eq.(ii), we get $$\begin{align} & { }^n C_4=\dfrac{n !}{(n-4) ! 4 !}=35 \\ & \Rightarrow \dfrac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}=35 \times 24=840 \\ & \Rightarrow n(n-1)(n-2)(n-3)=840 \\ & \Rightarrow n(n-3)(n-1)(n-2)=840 \\ & \Rightarrow(n^2-3 n)(n^2-3 n+2)=840 \end{align}$$ Let $y=n^2-3 n$ then the above last equation becomes $$\begin{align} & y(y+2)=840 \\ & \Rightarrow y^2+2 y-840=0 \\ & \Rightarrow y^2+30 y-28 y-840=0 \\ & \Rightarrow y(y+30)-28(y+30)=0\\ &\Rightarrow(y-28)(y+30)=0\\ \Rightarrow \text{Either} y&=28, \text{or} y=-30\end{align}$$ When $y=28$ then $$\begin{align} & n^2-3 n=28 \\ & \Rightarrow n^2-3 n-28=0 \\ & \Rightarrow n^2-7 n-4 n-28=0 \\ & \Rightarrow n(n-7)+4(n-7)=0 \\ & \Rightarrow(n-7)(n+4)=0 \end{align}$$ $\Rightarrow$ Either $n=7$, or $n=-4$. But $n$ can not be negative, so $n=7$. $$\begin{align}\text{When} &y=-30 \text{then}\\ & n^2-3 n=-30 \\ & \Rightarrow n^2-3 n+30=0\end{align}$$ $\Rightarrow n=\dfrac{3 \pm \sqrt{111} i}{2}$ But $n$ can not be complex. hence the only value of $n=7$ and $r=4$.

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