Combination

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^{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 and 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)(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}

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