Question 3 and 4 Exercise 6.2

Solutions of Question 3 and 4 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.

Prove by Fundamental principle of counting $^n P_r=n(^{n-1} P_{r-1})$

We are given that: $$^n P_r=n({ }^{n-1} P_{r-1})$$ We are taking the right hand side of the equation $$\begin{align}n(^{n-1} P_{r-1})&=n \dfrac{(n-1) !}{((n-1)-(r-1)) !} \\ & =\dfrac{n(n-1) !}{(n-r) !}\\ &=\dfrac{n !}{(n-r) !}\\ &=^n P_r\end{align}$$

Prove by Fundamental principle of counting $^n P_r=^{n-1} P_r+r(^{n-1} P_{r-1})$

We are given: $$^n P_r=^{n-1} P_r+r({ }^{n-1} P_{r-1})$$ Taking R.H.S of the equation $$\begin{align}^{n-1} P_r+r(^{n-1} P_{r-1})&=\dfrac{(n-1) !}{(n-1-r) !}+r \dfrac{(n-1) !}{(n-1-(r-1)) !} \\ & =\dfrac{(n-1) !}{(n-r-1) !}+r \dfrac{(n-1) !}{(n-r) !} \\ & =\dfrac{(n-1) !}{(n-r-1) !}+r \dfrac{(n-1) !}{(n-r)(n-r-1) !} \\ & =\dfrac{(n-1) !}{(n-r-1) !}[1+\dfrac{r}{n-r}] \\ & =\dfrac{(n-1) !}{(n-r-1) !}[\dfrac{n-r+r}{n-r}] \\ & =\dfrac{n(n-1) !}{(n-r)(n-r-1) !}\\ &=\dfrac{n !}{(n-r) !} \\ & =^n P_r\end{align}$$ which is the desired L.H.S.

In how many ways can a police department arrange eight suspects in a line up?

The total number of suspects are eight sọ, $n=8$

The total number of arrangements are: $$\begin{align}^n P_r&=^8 P_8\\ &=\dfrac{8 !,}{(8-8) !}\\ &=8 !=40,320\end{align}$$

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