Question 1, Exercise 6.2

Solutions of Question 1 of Exercise 6.2 of Unit 06: Permutation and Combination. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Prove for $n \in N$: $\quad^nP_r=\dfrac{n!}{(n-r)!}$

Solution.

Let us have $n$ distinct objects and we want to arrange $r$ of them in some order.
For first object, we have $\dot{n}$-choices
for $2^{\text {nd }}$ object, we have $(n-1)$ choices
For $3^{\text {rd }}$ object, we have $(n-2)$ choices
For $r^{\text {th }}$ object, we have $(n-(r-1)$ ) choices
Hence no. of possible arrangements

$${ }^{n} P_{r}=n(n-1)(n-2) \ldots(n-(r-1))$$

Multiply and divide by $(n-r)(n-r-1) \ldots 3,2.1$
${ }^{n} P_{r}=\frac{n(n-1)(n-2)\ldots(n-(r-1))(n-r)(n-r-1)(n-r-2) \ldots 3.2 .1}{(n-r)(n-r-1)(n-r-2) \ldots 3.2 .1}$

$$=\dfrac{n!}{(n-r)!}$$

Prove for $n \in N$: $\quad^nP_n=^nP_{n-1}$

Solution.

$$\begin{align*} L.H.S&=^{n} p_{n}=\dfrac{n!}{(n-n)!}\\ &=\dfrac{n!}{0!}=n!\\ R.H.S &={ }^{n} P_{n-1}\\ &=\dfrac{n!}{(n-(n-1))!}\\ &=\dfrac{n!}{(n-n+1)!}\\ &=\dfrac{n!}{1!}=n!\because 1!=1\\ \Rightarrow\quad L.H.S &= R.H.S\end{align*}$$

Prove for $n \in N$: $\quad^nP_r=n\quad^{n-1}P_{r-1}$

Solution.

$$\begin{align*}R. H.S &=n^{n-1} P_{r-1}\\ & =n \dfrac{(n-1)!}{((n-1)-(r-1))!} \\ & =\dfrac{n(n-1)!}{\left(n-1-r+1\right)!}\\ &=\dfrac{n!}{(n-r)!}\\ &={ }^{n} P_{r}=\text { L.H.S } \end{align*}$$

Prove for $n \in N$: $\quad^nP_r=^{n-1}P_r+r\quad^{n-1}P_{r-1}$

Solution.

$$\begin{align*}R.H.S &={ }^{n-1} P_{r}+r^{n-1} P_{r-1}\\ & =\dfrac{(n-1)!}{((n-1)-r)!}+r \dfrac{(n-1)!}{((n-1)-(r-1))!} \\ & =\dfrac{(n-1)!}{(n-r-1)!}+\dfrac{r(n-1)!}{(n-r)!} \\ & =(n-1)!\left[\dfrac{1}{(n-r-1)!}+\dfrac{r}{(n-r)(n-r-1)!}\right] \\ & =(n-1)!\left[\dfrac{(n-r)+r}{(n-r)(n-r-1)}\right] \\ & =(n-1)!\left[\frac{n}{(n-r)(n-r-1)!}\right]\\ &=\dfrac{n(n-1)!}{(n-r)(n-r-1)!}\\ &=\dfrac{n!}{(n-r)!}\\ &={ }^{n} P_{r}=L.H.S\end{align*}$$

Prove for $n \in N$: $\quad^n P_n=2\cdot\, ^n P_{n-2}$

Solution.

$$\begin{align*}R.H.S =2 \cdot{ }^{n} P_{n-2}\\ & =2 \cdot \dfrac{n!}{(n-(n-2))!}\\ &=2 \cdot \dfrac{n!}{(n-n+2)!}\\ &=2 \cdot \dfrac{n!}{2!} \\ & = 2 \cdot \dfrac{n!}{ 2}=n!\quad \cdots (i)\\ L.H.S &={ }^{n} P_{n}\\ &=\dfrac{n!}{(n-n)!}\\ &=\dfrac{n!}{0!}=n!\quad \cdots(ii)\end{align*}$$ From (i) and (ii) $$L.H.S = R.H.S$$

Question 2>