Question 1 and 2 Exercise 6.2

Solutions of Question 1 and 2 of Exercise 6.2 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.

Evaluate $^6 P_6$

$$\begin{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &=6 !=720\end{align}$$

Evaluate $^{20} P_2$

$$\begin{align}^{20} P_2&=\dfrac{20 !}{(20-2) !}\\ &=\dfrac{20.19 .18 !}{18 !}\\ &=20 \times 19=380\end{align}$$

Evaluate $^{16} P_3$

$$\begin{align}^{16} P_3&=\dfrac{16 !}{(16-3) ! }\\ &=\dfrac{16 \cdot 15 \cdot 14 \cdot 13 !}{13 ! }\\ &=3360 \end{align}$$

Solve $^n P_5=56(^n P_3)$ for $n.$

We are given: $$\begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \dfrac{n !}{(n-5) !}&=56 \dfrac{n !}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3)(n-4)(n-5) !} \\ \Rightarrow(n-3)(n-4)&=56 \\ \Rightarrow n^2-7 n+12-56&=0 \\ \Rightarrow n^2-7 n-44&=0 \\ \Rightarrow n^2+4 n-11 n-44&=0 \\ \Rightarrow n(n+4)-11(n+4)&=0 \\ \Rightarrow(n-11)(n+4)&=0\\ n=11& \text{or}\quad n=-4\end{align}$$ But $n$ can not be negative, so $n=11$.

Solve $^n P_5=9(^{n-1} P_4)$ for $n.$

We are given: $$\begin{align} ^n P_5&=9(^{n-1} P_4) \\ \Rightarrow \dfrac{n !}{(n-5) !}\\ &=9 \dfrac{(n-1) !}{(n-1-4) !} \\ \Rightarrow \dfrac{n(n-1) !}{(n-5) !}\\ &=\dfrac{9(n-1) !}{(n-5) !} \\ \Rightarrow n&=9 \end{align}$$

Solve $n^2 P_2=600$ for $n$

We are given: $$\begin{align}n^2 P_2&=60 \\ \Rightarrow \dfrac{(n^2) !}{(n^2-2) !}&=600 \\ \Rightarrow \dfrac{(n^2)(n^2-1)(n^2-2) !}{(n^2-2) !}&=600 \\ \Rightarrow n^4-n^2-600&=0 \\ \Rightarrow n^4-25 n^2+24 n^2-600&=0 \\ \Rightarrow n^2(n^2-25)+24(n^2-25)&=0 \\ \Rightarrow(n^2-25)(n^2+24)=0\\ n^2-25=0\quad\text{or}\quad n^2+24&=0\\ n^2=25\quad\text{or}\quad n^2&=-24\\ n=\pm 5\quad\text{or}\quad n&=\sqrt{24}i\end{align}$$ but $n$ can not be negative nor it can be imaginary, therefore, $n=5$.

Question 3 & 4 >