Question 1(i-v), Exercise 6.3

Solutions of Question 1(i-v) of Exercise 6.3 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 $n\in N$: $\quad^nC_r=\dfrac{n!}{r!(n-r)!}$

Solution.

Let us have $n$ distinct objects and we want to take $r$ at a time where $0<r<n$.
Let $X$ be the total combinations.
These $r$ objects may be arranged in $r$ ! ways but all $r$ ! arrangement would be treated as one combination. i.e.,

$$\begin{array}{ll} & r!X={ }^{n} P_{r} \\ \Rightarrow & r!X=\frac{n!}{(n-r)!} \\ \Rightarrow & X=\frac{n!}{r!(n-r)!}={ }^{n}{C}_{r} \end{array}$$

Prove for $n\in N$: $n\cdot^{n-1}C_{r-1}=(n-r+1)\quad ^nC_{r-1}$

Solution.

$$\begin{align*}L.H.S &=n\cdot^{n-1}C_{r-1} \\ &=n \cdot \dfrac{(n-1)!}{(r-1)!((n-1)-(r-1))!}\\ &=\dfrac{n(n-1)!}{(r-1)!(n-r)!}\\ &=\dfrac{n!}{(r-1)!(n-r)!}\end{align*}$$

Multiply and divide by $(n-r+1)$ $$\begin{align*} & =(n-r+1) \frac{n!}{(r-1)!(n-r+1)(n-r)!} \\ & \quad \because(n-r)!(n-r+1)=(n-r+1)!\\ \Rightarrow \quad&=(n-r+1) \frac{n!}{(r-1)!(n-r+1)!}\\ & =(n-r+1) \frac{n!}{(r-1)!(n-(r-1))!}\\ &=(n+r+1)^{n} C_{r-1}=\text { R.H.S }\end{align*}$$

Prove for $n\in N$: $r\, ^nC_r=(n-r+1)^nC_{r-1}$

Solution.

$$\begin{align*} L.H.S &=r \cdot \dfrac{n!}{r!(n-r)!}\\ &=\gamma \cdot \dfrac{n!}{\gamma(r-1)!(n-r)!}\\ &=\dfrac{n!}{(r-1)!(n-r)!} \tag{r-1} \end{align*}$$ Multiply and divide by $(n-r+1)$ $$\begin{align*} & =(n-r+1) \frac{n!}{(r-1)!(n-r+1)(n-r)!} \\ & =(n-r+1) \frac{n!}{(r-1)!(n-r+1)!} \\ & =(n-r+1) \frac{n!}{(r-1)!(n-(r-1))!}\\ &=(n-r+1)^{n} C_{r-1}=\text { R.H.S } \end{align*}$$

Prove for $n\in N$: $\quad^{n-1}C_{r-1}+^{n-1}C_{r}=^nC_r$

Solution.

$$\begin{align*} L.H.S &=\dfrac{(n-1)!}{(r-1)!((n-1)-(r-1))!}+\dfrac{(n-1)!}{r!(n-1-r)!}\\ &=\dfrac{(n-1)!}{(r-1)!(n-r)!}+\dfrac{(n-1)!}{r!(n-r-1)!}\\ \because r!&=r(r-1)!\&(n-r)!=(n-r)(n-r-1)!\\ &=\dfrac{(n-1)!}{(r-1)!(n-r)(n-r-1)!}+\dfrac{(n-1)!}{r(r-1)!(n-r-1)!}\\ &=\dfrac{(n-1)!r+(n-1)!(n-r)}{r(r-1)!(n-r)(n-r-1)!}\\ &=\dfrac{(n-1)!(\gamma+n-\gamma)}{r!(n-r)!}\\ \because r(r-1)&=r!\&(n-r)(n-r-1)!=(n-r)!\\ &=\dfrac{n(n-1)!}{r!(n-r)!}\\ &=\dfrac{n!}{r!(n-r)!}\\ &={ }^{n} C_{r}=\text { R.H.S } \quad \because n(n-1)!=n!\end{align*}$$

Prove for $n\in N$: $\quad^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_r$

Solution.

$$\begin{align*}L.H.S &=\dfrac{n!}{r!(n-r)!}+\dfrac{n!}{(r-1)!(n-(r-1))!}\\ &=\dfrac{n!}{r!(n-r)!}+\dfrac{n!}{(r-1)!(n-r+1)!}\\ &=\dfrac{n!}{r(r-1)!(n-r)!}+\dfrac{n!}{(r-1)!(n-r+1)(n-r)!}\\ &\quad \because r!=r(r-1)!\\ &\quad \because (n-r+1)!=(n-r+1)(n-r)!\\ &=\dfrac{n!(n-r+1)+n!(r)}{r(n-r)!(n-r+1)(n-r)!}\\ &=\dfrac{n!(n+1)}{r!(n-r+1)!}\\ &=\dfrac{(n+1)!}{r!(n-r+1)!}\\ &=\,\,^{n+1}C_r=R.H.S.\end{align*}$$

Question 1(vi-x)>