EXERCISE 17.1

QUESTION 1

Evaluate the following :

(i) $^{14}C_{3}$

Sol :

$\Rightarrow ~^{14}C_{3}=\dfrac{14}{3} \times \dfrac{13}{2} \times \dfrac{12}{1} \times~^{11}C_{0}$ $\left[ \begin{array}{c}{\because~^{n}C_{r}=\dfrac{n}{r}\times~^{n-1}C_{r-1} ]}\end{array}\right]$

$\Rightarrow~^{14}C_{3}=364$ $\left[\because~^{n}C_{0}=1\right]$

(ii) $^{12}C_{10}$

Sol :

$\Rightarrow ~^{12}C_{10}=~^{12}C_{2}$ $\left[ \begin{array}{c}{\because~^{n}C_{r}=~^{n}C_{n-r} ]}\end{array}\right.$

$\Rightarrow~^{12}C_{10}=\dfrac{12}{2} \times \dfrac{11}{1} \times~^{10}C_{0}$ $\left[\because~^{n} C_{r}=\dfrac{n}{r}\times~^{n-1}C_{r-1}\right]$

$\Rightarrow~^{12}C_{10}=\dfrac{12}{2} \times \dfrac{11}{1} \times 1$ $\left[\because~^{n} C_{0}=1\right]$

$\Rightarrow~^{12}C_{10}=66$

(iii) $^{35}C_{35}$

Sol :

$\Rightarrow ^{35}C_{35}=1$ $\left[\because~^{n}C_{n}=1\right]$

(iv) $^{n+1}C_{n}$

Sol :

$=\dfrac{(n+1) !}{(n !)(n+1-n) !}$ $\left(\because~^{n}C_{r}=\dfrac{n !}{r !(n-r) !}\right)$

$=\dfrac{(n+1) \times n !}{n ! \times 1 !}$

$=n+1$

(v) $\sum_{r=1}^{5}~^{5}C_{r}$

Sol :

$=~^{5}C_{1}+~^{5}C_{2}+~^{5}C_{3}+~^{5}C_{4}+~^{5}C_{5}$

$=\dfrac{5 !}{1 ! ~4 !}+\dfrac{5 !}{2 !~ 3 !}+\dfrac{5}{3 ! ~2 !}+\dfrac{5 !}{4 ! ~1 !}+\dfrac{5 !}{5 ! ~0 !}$ $\left(\because~^{n}C_{r}=\dfrac{n !}{r !(n-r) !}\right)$

$=5+\dfrac{5 \times 4}{2}+\dfrac{5 \times 4}{2}+5+1$

$=5+10+10+5+1$

= 31

QUESTION 2

If $^{n}C_{12}=~^{n}C_{5,}$ find the value of n

Sol :

Given : $^{n}C_{12}=~^{n}C_{5}$

$\Rightarrow n=12+5=17$ $\left[ \because~^nC_x=~^nC_y\Rightarrow~x=y \text{ or } , n=x+y\right]$

QUESTION 3

If $^{n}C_{4}=~^{n}C_{6},$ find $^{12}C_{n}$

Sol :

$^{n}C_{4}=^{n}C_{6}$

$\Rightarrow~n=6+4=10$
$\left[\because~^{n}C_{x}=~^{n}C_{y} \Rightarrow~x=y\text{ or } n=x+y\right]$

Now, $^{12}C_{10}=~^12C_{2}$ $\left[\because~^{n}C_{r}=~^{n}C_{n-r}\right]$

$=\dfrac{12}{2} \times \dfrac{11}{1} \times~^{10}C_{0}$ $\left[\because~^{n}C_{r}=\dfrac{n}{r}~^{n-1}C_{r-1}\right]$

= 66 $\left[\because~^{n}C_{0}=1\right]$

QUESTION 4

If $^{n}C_{10}=~^{n}C_{12},$ find $^{23}C_{n}$

Sol :

We have,

$^{n}C_{10}=~^{n}C_{12}$

$\Rightarrow~n=12+10=22$ $\left[ \begin{array}{ll}{\because~^{n}C_{x}=~^{n}C_{y}} & {\Rightarrow~x=y \text { or, } n=x+y ]}\end{array}\right.$

Now $~^{23} \mathrm{C}_{22}=~^{23} \mathrm{C}_{1}$ $\left[\because~^{n}C_{r}=~^{n}C_{n-r}\right]$

$=\dfrac{23}{1} \times~^{22}C_{0}$ $\left[\because~^{n}C_{r}=\dfrac{n}{r} ~^n-1C_{r-1}\right]$

= 23 $\left[\because~^{n}C_{0}=1\right]$

QUESTION 5

If $~^{24}C_{X}=~^{24}C_{2 x+3},$ find x

Sol :

$24=x+2 x+3$ $\left[\because~^{n}C_{x}=~^{n}C_{y} \Rightarrow~x=y \text{or}n=x+y\right]$

$24=3 x+3$

$3 x=21$

$x=7$

QUESTION 6

If $~^{18}C_{X}=~^{18}C_{X+2},$ find x

Sol :

$18=x+x+2$ $\left[\because~^{n}C_{x}=~^{n}C_{y} \Rightarrow x=y \text{ or } n=x+y\right]$

$2 x=16$

$x=8$

QUESTION 7

If $~^{15}C_{3 r}=~^{15}C_{r+3,}$ find r

Sol :

$15=3 r+r+3$ $\left[\because~^{n}C_{x}=~^{n}C_{y} \Rightarrow x=y \text{ or } n=x+y\right]$

$15=4 r+3$

$4 r=12$

$r=3$

QUESTION 8

If $~^{8}C_{r}-~^{7}C_{3}=~^{7}C_{2},$ find r

Sol :

$^{8}C_{r}=~^{7}C_{2}+~^{7}C_{3}$

$^{8}C_{r}=~^{8}C_{3}$ $\left[\because~^{n}C_{r}+~^{n}C_{r-1}=~^{n+1}C_{r} ; r \leq n\right]$

$r=3$ $\left[ \begin{array}{c}{\because~^{n}C_{x}=~^{n}C_{y} \Rightarrow x=y \text { or } n=x+y ]}\end{array}\right.$

And $r+3=8$

$\Rightarrow~r=5$

QUESTION 9

If $^{15}C_{r} :~^{15}C_{r-1}=11 : 5,$ find r

Sol :

We have,
$\dfrac{15}{15}C_{r-1}=\dfrac{11}{5}$

$\Rightarrow~\dfrac{15-r+1}{r}=\dfrac{11}{5}$ $\left[\because \dfrac{^{n}C_{r}}{^{n}C_{r-1}}=\dfrac{n-r+1}{r}\right]$

$\Rightarrow~75-5 r+5=11 r$

$\Rightarrow~16 r=80$

$\Rightarrow~r=5$

QUESTION 10

If $^{n+2}C_{8} :~^{n-2}P_{4}=57 : 16,$ find n

Sol :

We have

$\Rightarrow \quad \dfrac{^{n+2}C_{8}}{^{n-2}P_{4}}=\dfrac{57}{16}$

$\Rightarrow \dfrac{(n+2) !}{8 !(n-6) !} \times \dfrac{(n-6) !}{(n-2) !}=\dfrac{57}{16}$

$\Rightarrow \dfrac{(n+2)(n+1) n(n-1)(n-2) !}{8 !} \times \dfrac{1}{(n-2) !}=\dfrac{57}{16}$

$\Rightarrow(n+2)(n+1) n(n-1)=\dfrac{57}{16} \times 8 !$

$=\dfrac{19 \times 3}{16} \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$\Rightarrow(n+2)(n+1) n(n-1)=143640$

$\Rightarrow(n-1) n(n+1)(n+2)=19 \times 3 \times 7 \times 6 \times 5 \times 4 \times 3$

$\Rightarrow(n-1) n(n+1)(n+2)=19 \times(3 \times 7) \times(6 \times 3) \times(4 \times 5)$

$\Rightarrow(n-1) n(n+1)(n+2)=18 \times 19 \times 20 \times 21$

$\Rightarrow~n-1=18$

$\Rightarrow~n=19$

QUESTION 11

If $^{28}C_{2 r} :~^{24}C_{2 r-4}=225 : 11$ , find r

Sol :

We have

$^{28}C_{2 r} :~^{24}C_{2 r-4}=225 : 11$

$\Rightarrow \dfrac{~^{28}C_{2 r}}{^{24}C_{2 r-4}}=\dfrac{225}{11}$

$\Rightarrow \quad \dfrac{28 !}{2 r !(28-2 r) !} \times \dfrac{(2 r-4) !(28-2 r) !}{24 !}=\dfrac{225}{11}$

$\Rightarrow \dfrac{28 \times 27 \times 26 \times 25}{2 r(2 r-1)(2 r-2)(2 r-3)}=\dfrac{225}{11}$

$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=\dfrac{28 \times 27 \times 26 \times 25 \times 11}{225}$

$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=28 \times 3 \times 26 \times 11$

$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=4 \times 7 \times 3 \times 13 \times 2 \times 11$

$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=(2 \times 7) \times 13 \times(3 \times 4) \times 11$

$\Rightarrow 2 r(2 r-1)(2 r-2)(2 r-3)=14 \times 13 \times 12 \times 11$

$\Rightarrow 2 r=14$

$\Rightarrow r=7$

QUESTION 12

If $^{n}C_{4},~^{n}C_{5}$ and $^{n}C_{6}$ are in A.P. then find n

Sol :

Since $^{n} C_{4},~^{n}C_{5}$ and $^{n}C_{6}$ are in AP.

[ if a ,b ,c are in A.P then 2b=a+c ]

$\Rightarrow~2~^{n}C_{5}=~^{n}C_{4}+~^{n}C_{6}$

$\Rightarrow~2 \times \dfrac{n !}{5 !(n-5) !}=\dfrac{n !}{4 !(n-4) !}+\dfrac{n !}{6 !(n-6) !}$

$\Rightarrow \dfrac{2}{5 \times 4 !(n-5)(n-6) !}=\dfrac{1}{4 !(n-5)(n-4)(n-6) !}+\dfrac{1}{6 \times 5 \times 4 !(n-6) !}$

$\Rightarrow \dfrac{2}{5(n-5)}=\dfrac{1}{(n-5)(n-4)}+\dfrac{1}{30}$

$\Rightarrow \dfrac{2}{5(n-5)}-\dfrac{1}{(n-5)(n-4)}=\dfrac{1}{30}$

$\Rightarrow \dfrac{2 n-8-5}{5(n-5)(n-4)}=\dfrac{1}{30}$

$\Rightarrow 12 n-78=n^{2}-9 n+20$

$\Rightarrow n^{2}-21 n+98=0$

$\Rightarrow(n-7)(n-14)=0$

$\therefore~n=7$ and 14

QUESTION 13

If $^{2 n}C_{3} :~^{n}C_{2}=44 : 3,$ find n

Sol :

Given : $^{2 n}C_{3} :~^{n}C_{2}=44 : 3$

$\dfrac{^{2 n}C_{3}}{^{n}C_{2}}=\dfrac{44}{3}$

$\Rightarrow \dfrac{2 n(2 n-1)(2 n-2)}{3 n(n-1)}=\dfrac{44}{3}$

$\Rightarrow(2 n-1)(2 n-2)=22(n-1)$

$\Rightarrow 4 n^{2}-6 n+2=22 n-22$

$\Rightarrow 4 n^{2}-28 n+24=0$

$\Rightarrow n^{2}-7 n+6=0$

$\Rightarrow n^{2}-6 n-n+6=0$

$\Rightarrow n(n-6)-1(n-6)=0$

$\Rightarrow(n-1)(n-6)=0$

$\Rightarrow n=1$ or, $n=6$

Now, $n=1 \Rightarrow~^2 \mathrm C_{3} : ^2\mathrm C_{2}=44 : 3$

But this is not possible

$\therefore n=6$

QUESTION 14

If $^{16}C_{r}=~^{16}C_{r+2},$ find $^{r}C_{4}$

Sol :

$^{16}C_{r}=~^{16}C_{r+2},$

$16=r+r+2$ $\left[^{n}C_{x}=~^{n}C_{y} \Rightarrow x=y \text{ or } x+y=n\right]$

$\Rightarrow 2 r+2=16$

$\Rightarrow 2 r=14$

$\Rightarrow r=7$

Now , $^{r} C_{4}=^{7} C_{4}$

$\Rightarrow ~^7C_{4}=~^{7}C_{3}$ $\left[\because~^nC_r=~^nC_r\right]$

$\Rightarrow~^{7} C_{3}=\dfrac{7}{3} \times \dfrac{6}{2} \times \dfrac{5}{1} \times~^{4} C_{0}$ $\left[\because~^{n}C_{r}=\dfrac{n}{r} \times~^{n-1}C_{r-1}\right]$

$\Rightarrow~^{7}C_{4}=35$ $\left[\because~^{n}C_{0}=1\right]$

QUESTION 15

If $a=~^{m}C_{2},$ then find the value of $^{a}C_{2}$

Sol :

$^{\alpha} C_{2}=\dfrac{\alpha}{2} \times \dfrac{(\alpha-1)}{1} \times~^{\alpha} C_{0}$ $\left[\because~^{n}C_{r}=\dfrac{n}{r} \times~^{n-1}C_{r-1}\right]$

$=\dfrac{1}{2} \alpha(\alpha-1)$ $\left[\because~^{n}C_{0}=1\right]$

$=\dfrac{1}{2}\left[~^{m}C_{2}\left(~^{m}C_{2}-1\right)\right]$

$=\dfrac{1}{2}\left[\dfrac{m !}{2 !(m-2) !}\left(\dfrac{m !}{2 !(m-2) !}-1\right)\right]$

$=\dfrac{1}{2}\left[\dfrac{m(m-1)}{2}\left(\dfrac{m(m-1)}{2}-1\right)\right]$

$=\dfrac{1}{2}\left[\dfrac{m(m-1)}{2}\left(\dfrac{m(m-1)-2}{2}\right)\right]$

$=\dfrac{1}{8}[m(m-1)\{m(m-1)-2\}]$

$=\dfrac{m(m-1)\left(m^{2}-m-2\right)}{2 \times 2 \times 2}$

$=\dfrac{m(m-1)(m+1)(m-2)}{8}$

$=\dfrac{m(m-1)(m+1)(m-2)}{4 \times 2}$

multiplying with 3 to numerator and denominator to make 4 :

$=\dfrac{m(m+1) m(m-1)(m-2)}{4.3 .2 .1}$

$=\dfrac{3(m+1) m(m-1)(m-2)}{4 !}$

$=\dfrac{3(m+1)(m)(m-1)(m-2)}{4 !}\times\dfrac{(m-3)!}{(m-3)!}$

$=\dfrac{3(m+1)(m)(m-1)(m-2)(m-3)!}{4 !(m-3)!}$

$=3 ~^{m+1} C_{4}$ $\left(\because~^{n} C_{r}=\dfrac{n !}{r !(n-r) !}\right)$

QUESTION 16

Prove that the product of 2n consecutive negative integers is divisible by $(2 n) !$

Sol :

Let 2n negative integers be $(-r),(-r-1),(-r-2), \ldots, \ldots,(-r-2 n+1)$

Then, product $=(-1)^{2 n}(r)(r+1)(r+2), \ldots, \ldots(r+2 n-1)$

$=\dfrac{(r-1) !(r)(r+1)(r+2) \ldots \ldots(r+2 n-1)}{(r-1) !}$

$=\dfrac{(r+2 n-1) !}{(r-1) !}$

$=\dfrac{(r+2 n-1) !}{(r-1) !(2 n) !} \times(2 n) !$

$=r+2 n-1 C_{2 n} \times(2 n) !$

This is divisible by $(2 n) !$

[Alternate method]

Product $=[(2 n+1)(2 n+3)(2 n+5) \dots(2 n+r)]$

$=\dfrac{(2 n) ![(2 n+1)(2 n+3) \dots(2 n+r)]}{(2 n) !}$

$=\dfrac{(2 n)[(2 n-1)(2 n-2) \dots 4.2(2 n+1)(2 n+3)]}{(2 n) !}$

$=\dfrac{(2 n+r) !}{(2 n) !}$

Hence $r=2 n$

$=\dfrac{(2 n+2 n) !}{2 n}$

$=\dfrac{(4 n) !}{(2 n) !}$

$=(2 n) !$

This is divisible by $(2 n) !$

QUESTION 17

For all positive integers n , show that $^{2 n}C_{n}+~^{2 n}C_{n-1}=\dfrac{1}{2}\left(^{2 n+2}C_{n+1}\right)$

Sol :

LHS $=^{2 n}C_{n}+~^{2 n}C_{n-1}$

$=\dfrac{(2 n) !}{n ! n !}+\dfrac{(2 n) !}{(n-1) !(2 n-n+1) !}$

$=\dfrac{(2 n) !}{n ! n !}+\dfrac{(2 n) !}{(n-1) !(n+1) !}$

$=\dfrac{(2 n) !}{n(n-1) ! n !}+\dfrac{(2 n) !}{(n-1) !(n+1) n !}$

$=\dfrac{(2 n) !}{n !(n-1) !}\left[\dfrac{1}{n}+\dfrac{1}{n+1}\right]$

$=\dfrac{(2 n) !}{n !(n-1) !}\left[\dfrac{2 n+1}{n(n+1)}\right]$

$=\dfrac{(2 n+1) !}{n !(n+1) !}$

RHS $=\dfrac{1}{2}\times ~^{2 n+2}C_{n+1}$

$=\dfrac{1}{2}\left[\dfrac{(2 n+2) !}{(n+1) !(2 n+2-n-1) !}\right]$

$=\dfrac{1}{2}\left[\dfrac{(2 n+2) !}{(n+1) !(n+1) !}\right]$

$=\dfrac{1}{2}\left[\dfrac{(2 n+2)(2 n+1) !}{(n+1) n !(n+1) !}\right]$

$=\dfrac{1}{2}\left[\dfrac{2(n+1)(2 n+1) !}{(n+1) n !(n+1) !}\right]$

$=\dfrac{(2 n+1) !}{n !(n+1) !}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

QUESTION 18

Prove that: $^{4 n} C_{2 n} :~^{2 n} C_{n}=[1 \cdot 3 \cdot 5 \ldots(4 n-1)] :[1 \cdot 3 \cdot 5 \ldots(2 n-1)]^{2}$

Sol :

$\dfrac{4 n C_{2 n}}{2 n C_{n}}=\dfrac{1.3 .5 \ldots(4 n-1)}{[1.3 .5 . . .(2 n-1)]^{2}}$

$\mathrm{LHS}=\dfrac{^{4 n}C_{2 n}}{^{2 n}C_{n}}$

$=\dfrac{(4 n) !}{(2 n) !(2 n) !} \times \dfrac{n ! \times n !}{(2 n) !}$

$=\dfrac{\left[4n\times (4n-1)\times (4n-2)\dots 3\times 2\times 1\right]\times (n!)^2}{\left[2n\times (2n-1)\times (2n-2)\dots 3\times 2\times 1\right]^2\times (2n)!}$

$=\dfrac{\left[1\times 3\times 5 \dots (4n-1)\right] \left[2\times 4\times 6 \dots (4n)\right]\times (n!)^2}{\left[1\times 3\times 5 \dots (2n-1) \right]^2~\left[2\times 4\times 6\dots 2n\right]^2 \times (2n)!}$

$=\dfrac{\left[1\times 3\times 5 \dots (4n-1)\right]\times (2)^2n\times \left[1\times 2\times 3 \dots 2n\right](n!)^2}{\left[1\times 3\times 5 \dots (2n-1)\right]^2\times (2)^2n\times \left[1\times 2\times 3 \dots n^2\right]\times (2n)!}$

$=\dfrac{[1 \times 3 \times 5 \ldots \ldots .(4 n-1)](2 n) ![n !]^{2}}{[1 \times 3 \times 5 \times \ldots \ldots \ldots(2 n-1)]^{2}[n !]^{2}(2 n) !}$

$=\dfrac{[1 \times 3 \times 5 \ldots \ldots .(4 n-1)]}{[1 \times 3 \times 5 \times \ldots \ldots \ldots(2 n-1)]^{2}}$ $=\mathrm{RHS}$

Hence , proved

QUESTION 19

Evaluate $^{20} C_{5}+\sum_{r=2}^{5} ~^{25-r}C_{4}$

Sol :

$=^{20} C_{5}+~^{23} C_{4}+~^{22} C_{4}+~^{21} C_{4}+~^{20} C_{4}$

$=\left(^{20} C_{4}+~^{20} C_{5}\right)+~^{21} C_{4}+~^{22} C_{4}+~^{23} C_{4}$

$=~^{21} C_{5}+~^{21} C_{4}+~^{2} C_{4}+~^{23} C_{4}$ $\left[\because~^{n} C_{r-1}+~^{n} C_{r}=~^{n+1} C_{r}\right]$

$=\left(^{21} C_{4}+~^{21} C_{5}\right)+~^{22} C_{4}+~^{23} C_{4}$

$=~^{22} C_{5}+^{22} C_{4}+^{23} C_{4}$ $\left[\because~^{n} C_{r-1}+~^{n} C_{r}=~^{n+1} C_{r}\right]$

$=(~^{22} C_{5}+^{22} C_{4}+^{23} C_{4})$

$=^{23} C_{5}+^{23} C_{4}$ $\left[\because~^{n} C_{r-1}+~^{n} C_{r}=~^{n+1} C_{r}\right]$

$=~^{24} C_{5}$ $\left[\because~^{n} C_{r-1}+~^{n} C_{r}=~^{n+1} C_{r}\right]$

$=\dfrac{24 !}{19 ! 5 !}$

$=\dfrac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1}$

= 42504

QUESTION 20

Let r and n be positive integers such that $1 \leq r \leq n$ . Then prove the following

(i) $\dfrac{^{n} C_{r}}{^{n} C_{r-1}}=\dfrac{n-r+1}{r}$

Sol :

LHS $=\dfrac{^{n} C_{r}}{^{n} C_{r-1}}$

$=\dfrac{n !}{r !(n-r) !} \times \dfrac{(r-1) !(n-r+1) !}{n !}$

$=\dfrac{(n-r+1)(n-r) !(r-1) !}{r(r-1) !(n-r) !}$

$=\dfrac{n-r+1}{r}=\mathrm{RHS}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

(ii) $n \times ~^{n-1} C_{r-1}=(n-r+1)~^{n}C_{r-1}$

Sol :

$\mathrm{LHS}=n \times~^{n-1} C_{r-1}$

$=\dfrac{n(n-1) !}{(r-1) !(n-1-r+1) !}$

$=\dfrac{n !}{(r-1) !(n-r) !}$

$\mathrm{RHS}=(n-r+1)~^{n} C_{r}$

$=(n-r+1) \times \dfrac{n !}{(r-1) !(n-r+1) !}$

$=(n-r+1) \times \dfrac{n !}{(r-1) !(n-r+1)(n-r) !}$

$=\dfrac{n !}{(r-1) !(n-r) !}$

$\therefore \mathrm{L H S}=\mathrm{R H S}$

(iii) $\dfrac{^{n} C_{r}}{^{n-1} C_{r-1}}=\dfrac{n}{r}$

Sol :

$\mathrm{LHS}=\dfrac{^{n} C_{r}}{^{n-1}C_{r-1}}$

$=\dfrac{n !}{r !(n-r) !} \times \dfrac{(r-1) !(n-1-r+1) !}{(n-1) !}$

$=\dfrac{n(n-1) !}{r(r-1) !(n-r) !} \times \dfrac{(r-1) !(n-r) !}{(n-1) !}$

$=\dfrac{n}{r}=\mathrm{RHS}$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

(iv) $^{n} C_{r}+2 \times~^{n} C_{r-1}+~^{n} C_{r-2}=~^{n+2} C_{r}$

Sol :

$\mathrm{LHS}=^{n}C_{r}+2 \times ~^{n}C_{r-1}+~^{n}C_{r-2}=~^{n+2} C_{r}$

$=\left(^{n}C_{r}+~^{n} C_{r-1}\right)+\left(^{n} C_{r-1}+~^{n} C_{r-2}\right)$

$=~^{n+1}C_{r}+~^{n+1} C_{r-1}$ $\left[\because~^n{C}_{r}+~^nC_{r-1}=~^{n+1}C_{r}\right]$

$=~^{n+2}C_{r}$ $\left[\because~^nC_{r}+~{^n}C_{r-1}=~^{n+1}C_{r}\right]$

= R H S

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Combination

EXERCISE 17.1

Leave a Comment Cancel Reply