\displaystyle \textbf{Question 1: }\text{Evaluate the following:}
\displaystyle \text{i) }{}^{14}C_3\qquad\text{ii) }{}^{12}C_{10}\qquad\text{iii) }{}^{35}C_{35}\qquad\text{iv) }{}^{n+1}C_n
\displaystyle \text{v) }\sum\limits_{r=1}^{5}{}^{5}C_r
\displaystyle \text{Answer:}
\displaystyle \text{i) }{}^{14}C_3=\frac{14!}{3!(14-3)!}
\displaystyle =\frac{14!}{3!11!}
\displaystyle =\frac{14\times13\times12}{3\times2\times1}
\displaystyle =364
\displaystyle \text{ii) }{}^{12}C_{10}=\frac{12!}{10!(12-10)!}
\displaystyle =\frac{12!}{10!2!}
\displaystyle =\frac{12\times11}{2}
\displaystyle =66
\displaystyle \text{iii) }{}^{35}C_{35}=\frac{35!}{35!(35-35)!}
\displaystyle =\frac{35!}{35!0!}
\displaystyle =1
\displaystyle \text{iv) }{}^{n+1}C_n=\frac{(n+1)!}{n!(n+1-n)!}
\displaystyle =\frac{(n+1)!}{n!1!}
\displaystyle =n+1
\displaystyle \text{v) }\sum\limits_{r=1}^{5}{}^{5}C_r
\displaystyle =\sum\limits_{r=0}^{5}{}^{5}C_r-{}^{5}C_0
\displaystyle =2^5-1
\displaystyle =32-1
\displaystyle =31
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }{}^{n}C_{12}={}^{n}C_5,\text{ find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{n}C_{12}={}^{n}C_5
\displaystyle \text{We know that if }{}^{n}C_p={}^{n}C_q\text{ and }p\ne q,\text{ then }p+q=n.
\displaystyle \therefore 12+5=n
\displaystyle \therefore n=17
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }{}^{n}C_4={}^{n}C_6,\text{ find }{}^{12}C_n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{n}C_4={}^{n}C_6
\displaystyle \text{Since }4\ne6,\text{ we use the property }{}^{n}C_r={}^{n}C_{n-r}.
\displaystyle \therefore 4+6=n
\displaystyle \therefore n=10
\displaystyle {}^{12}C_n={}^{12}C_{10}
\displaystyle =\frac{12!}{10!2!}
\displaystyle =\frac{12\times11}{2}
\displaystyle =66
\displaystyle \therefore {}^{12}C_n=66.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }{}^{n}C_{10}={}^{n}C_{12},\text{ find }{}^{23}C_n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{n}C_{10}={}^{n}C_{12}
\displaystyle \text{Since }10\ne12,\text{ we use the property }{}^{n}C_r={}^{n}C_{n-r}.
\displaystyle \therefore 10+12=n
\displaystyle \therefore n=22
\displaystyle {}^{23}C_n={}^{23}C_{22}
\displaystyle =\frac{23!}{22!1!}
\displaystyle =23
\displaystyle \therefore {}^{23}C_n=23.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }{}^{24}C_x={}^{24}C_{2x+3},\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{24}C_x={}^{24}C_{2x+3}
\displaystyle \text{For }{}^{n}C_p={}^{n}C_q,\text{ either }p=q\text{ or }p+q=n.
\displaystyle \text{Case 1: }x=2x+3
\displaystyle \Rightarrow x=-3
\displaystyle \text{This value is not admissible since the lower index cannot be negative.}
\displaystyle \text{Case 2: }x+(2x+3)=24
\displaystyle \Rightarrow 3x+3=24
\displaystyle \Rightarrow 3x=21
\displaystyle \Rightarrow x=7
\displaystyle \therefore x=7.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }{}^{18}C_x={}^{18}C_{x+2},\text{ find }x.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{18}C_x={}^{18}C_{x+2}
\displaystyle \text{For }{}^{n}C_p={}^{n}C_q,\text{ either }p=q\text{ or }p+q=n.
\displaystyle \text{Case 1: }x=x+2
\displaystyle \text{This is not possible.}
\displaystyle \text{Case 2: }x+(x+2)=18
\displaystyle \Rightarrow 2x+2=18
\displaystyle \Rightarrow 2x=16
\displaystyle \Rightarrow x=8
\displaystyle \therefore x=8.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }{}^{15}C_{3r}={}^{15}C_{r+3},\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{15}C_{3r}={}^{15}C_{r+3}
\displaystyle \text{For }{}^{n}C_p={}^{n}C_q,\text{ either }p=q\text{ or }p+q=n.
\displaystyle \text{Case 1: }3r=r+3
\displaystyle \Rightarrow 2r=3
\displaystyle \Rightarrow r=\frac{3}{2}
\displaystyle \text{This value is not admissible since the lower index must be an integer.}
\displaystyle \text{Case 2: }3r+(r+3)=15
\displaystyle \Rightarrow 4r=12
\displaystyle \Rightarrow r=3
\displaystyle \therefore r=3.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }{}^{8}C_r-{}^{7}C_3={}^{7}C_2,\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle {}^{8}C_r-{}^{7}C_3={}^{7}C_2
\displaystyle \Rightarrow {}^{8}C_r={}^{7}C_2+{}^{7}C_3
\displaystyle \text{Using Pascal's identity, }{}^{n}C_r+{}^{n}C_{r+1}={}^{n+1}C_{r+1},
\displaystyle \Rightarrow {}^{8}C_r={}^{8}C_3
\displaystyle \therefore r=3\text{ or }8-3=5
\displaystyle \therefore r=3\text{ or }5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }{}^{15}C_r:{}^{15}C_{r-1}=11:5,\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{15}C_r:{}^{15}C_{r-1}=11:5
\displaystyle \Rightarrow \frac{{}^{15}C_r}{{}^{15}C_{r-1}}=\frac{11}{5}
\displaystyle \Rightarrow \frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}}=\frac{11}{5}
\displaystyle \Rightarrow \frac{15!}{r!(15-r)!}\times\frac{(r-1)!(16-r)!}{15!}=\frac{11}{5}
\displaystyle \Rightarrow \frac{16-r}{r}=\frac{11}{5}
\displaystyle \Rightarrow 5(16-r)=11r
\displaystyle \Rightarrow 80-5r=11r
\displaystyle \Rightarrow 16r=80
\displaystyle \Rightarrow r=5
\displaystyle \therefore r=5.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }{}^{n+2}C_8:{}^{n-2}P_4=57:16,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{n+2}C_8:{}^{n-2}P_4=57:16
\displaystyle \Rightarrow \frac{{}^{n+2}C_8}{{}^{n-2}P_4}=\frac{57}{16}
\displaystyle \Rightarrow \frac{\frac{(n+2)!}{8!(n-6)!}}{\frac{(n-2)!}{(n-6)!}}=\frac{57}{16}
\displaystyle \Rightarrow \frac{(n+2)!}{8!(n-6)!}\times\frac{(n-6)!}{(n-2)!}=\frac{57}{16}
\displaystyle \Rightarrow \frac{(n-1)n(n+1)(n+2)}{8!}=\frac{57}{16}
\displaystyle \Rightarrow (n-1)n(n+1)(n+2)=\frac{57\times8!}{16}
\displaystyle =57\times2520
\displaystyle =143640
\displaystyle =18\times19\times20\times21
\displaystyle \therefore n-1=18
\displaystyle \Rightarrow n=19
\displaystyle \therefore n=19.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }{}^{28}C_{2r}:{}^{24}C_{2r-4}=225:11,\text{ find }r.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{28}C_{2r}:{}^{24}C_{2r-4}=225:11
\displaystyle \Rightarrow \frac{{}^{28}C_{2r}}{{}^{24}C_{2r-4}}=\frac{225}{11}
\displaystyle \Rightarrow \frac{\frac{28!}{(2r)!(28-2r)!}}{\frac{24!}{(2r-4)!(28-2r)!}}=\frac{225}{11}
\displaystyle \Rightarrow \frac{28!}{(2r)!(28-2r)!}\times\frac{(2r-4)!(28-2r)!}{24!}=\frac{225}{11}
\displaystyle \Rightarrow \frac{28\times27\times26\times25}{(2r)(2r-1)(2r-2)(2r-3)}=\frac{225}{11}
\displaystyle \Rightarrow (2r)(2r-1)(2r-2)(2r-3)
\displaystyle =\frac{11\times28\times27\times26\times25}{225}
\displaystyle =14\times13\times12\times11
\displaystyle \Rightarrow (2r)(2r-1)(2r-2)(2r-3)=14\times13\times12\times11
\displaystyle \Rightarrow 2r=14
\displaystyle \Rightarrow r=7
\displaystyle \therefore r=7.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }{}^{n}C_4,\;{}^{n}C_5\text{ and }{}^{n}C_6\text{ are in A.P., find }n.
\displaystyle \text{Answer:}
\displaystyle \text{Since }{}^{n}C_4,\;{}^{n}C_5\text{ and }{}^{n}C_6\text{ are in A.P.,}
\displaystyle 2{}^{n}C_5={}^{n}C_4+{}^{n}C_6
\displaystyle \text{Dividing throughout by }{}^{n}C_5,
\displaystyle 2=\frac{{}^{n}C_4}{{}^{n}C_5}+\frac{{}^{n}C_6}{{}^{n}C_5}
\displaystyle \Rightarrow 2=\frac{5}{n-4}+\frac{n-5}{6}
\displaystyle \Rightarrow 12(n-4)=30+(n-5)(n-4)
\displaystyle \Rightarrow 12n-48=30+n^2-9n+20
\displaystyle \Rightarrow n^2-21n+98=0
\displaystyle \Rightarrow (n-7)(n-14)=0
\displaystyle \Rightarrow n=7\text{ or }n=14
\displaystyle \therefore n=7\text{ or }14.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }{}^{2n}C_3:{}^{n}C_2=44:3,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{2n}C_3:{}^{n}C_2=44:3
\displaystyle \Rightarrow \frac{{}^{2n}C_3}{{}^{n}C_2}=\frac{44}{3}
\displaystyle \Rightarrow \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{2!(n-2)!}}=\frac{44}{3}
\displaystyle \Rightarrow \frac{(2n)!}{3!(2n-3)!}\times\frac{2!(n-2)!}{n!}=\frac{44}{3}
\displaystyle \Rightarrow \frac{(2n)(2n-1)(2n-2)}{6}\times\frac{2}{n(n-1)}=\frac{44}{3}
\displaystyle \Rightarrow \frac{2(2n-1)(2n-2)}{3(n-1)}=\frac{44}{3}
\displaystyle \Rightarrow \frac{2(2n-1)\cdot2(n-1)}{n-1}=44
\displaystyle \Rightarrow 4(2n-1)=44
\displaystyle \Rightarrow 2n-1=11
\displaystyle \Rightarrow 2n=12
\displaystyle \Rightarrow n=6
\displaystyle \therefore n=6.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }{}^{16}C_r={}^{16}C_{r+2},\text{ find }{}^{r}C_4.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }{}^{16}C_r={}^{16}C_{r+2}
\displaystyle \text{For }{}^{n}C_p={}^{n}C_q,\text{ either }p=q\text{ or }p+q=n.
\displaystyle \text{Case 1: }r=r+2
\displaystyle \text{This is not possible.}
\displaystyle \text{Case 2: }r+(r+2)=16
\displaystyle \Rightarrow 2r+2=16
\displaystyle \Rightarrow 2r=14
\displaystyle \Rightarrow r=7
\displaystyle \therefore {}^{r}C_4={}^{7}C_4
\displaystyle =\frac{7!}{4!3!}
\displaystyle =\frac{7\times6\times5}{3\times2\times1}
\displaystyle =35
\displaystyle \therefore {}^{r}C_4=35.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\alpha={}^{m}C_2,\text{ find the value of }{}^{\alpha}C_2.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\alpha={}^{m}C_2
\displaystyle \therefore \alpha=\frac{m!}{2!(m-2)!}
\displaystyle =\frac{m(m-1)}{2}
\displaystyle {}^{\alpha}C_2=\frac{\alpha!}{2!(\alpha-2)!}
\displaystyle =\frac{\alpha(\alpha-1)}{2}
\displaystyle =\frac{1}{2}\times\frac{m(m-1)}{2}\left[\frac{m(m-1)}{2}-1\right]
\displaystyle =\frac{m(m-1)}{4}\left[\frac{m^2-m-2}{2}\right]
\displaystyle =\frac{m(m-1)(m-2)(m+1)}{8}
\displaystyle \therefore {}^{\alpha}C_2=\frac{m(m-1)(m-2)(m+1)}{8}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Prove that the product of }2n\text{ consecutive negative integers is}
\displaystyle \text{divisible by }(2n)!.
\displaystyle \text{Answer:}
\displaystyle \text{Let the }2n\text{ consecutive negative integers be}
\displaystyle -r,\;-(r+1),\;-(r+2),\ldots,-(r+2n-1),
\displaystyle \text{where }r\text{ is a positive integer.}
\displaystyle \text{Their product is}
\displaystyle (-r)(-r-1)(-r-2)\cdots(-r-2n+1)
\displaystyle =(-1)^{2n}r(r+1)(r+2)\cdots(r+2n-1)
\displaystyle =r(r+1)(r+2)\cdots(r+2n-1)
\displaystyle =\frac{(r+2n-1)!}{(r-1)!}
\displaystyle =\frac{(r+2n-1)!}{(2n)!(r-1)!}\times(2n)!
\displaystyle ={}^{r+2n-1}C_{2n}\times(2n)!
\displaystyle \text{Since }{}^{r+2n-1}C_{2n}\text{ is an integer, the product is divisible by }(2n)!.
\displaystyle \therefore \text{The product of }2n\text{ consecutive negative integers is divisible by }(2n)!.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{For all positive integers }n,\text{ show that}
\displaystyle {}^{2n}C_n+{}^{2n}C_{n-1}=\frac{1}{2}\left({}^{2n+2}C_{n+1}\right).
\displaystyle \text{Answer:}
\displaystyle \text{LHS}={}^{2n}C_n+{}^{2n}C_{n-1}
\displaystyle =\frac{(2n)!}{n!n!}+\frac{(2n)!}{(n-1)!(n+1)!}
\displaystyle =\frac{(2n)!}{(n-1)!n!}\left(\frac{1}{n}+\frac{1}{n+1}\right)
\displaystyle =\frac{(2n)!}{(n-1)!n!}\left(\frac{2n+1}{n(n+1)}\right)
\displaystyle =\frac{(2n+1)!}{n!(n+1)!}
\displaystyle \text{RHS}=\frac{1}{2}\left({}^{2n+2}C_{n+1}\right)
\displaystyle =\frac{1}{2}\left[\frac{(2n+2)!}{(n+1)!(n+1)!}\right]
\displaystyle =\frac{1}{2}\left[\frac{2(n+1)(2n+1)!}{(n+1)!(n+1)!}\right]
\displaystyle =\frac{(2n+1)!}{n!(n+1)!}
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore {}^{2n}C_n+{}^{2n}C_{n-1}=\frac{1}{2}\left({}^{2n+2}C_{n+1}\right).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove that}
\displaystyle {}^{4n}C_{2n}:{}^{2n}C_n=\big[1\cdot3\cdot5\cdots(4n-1)\big]:\big[1\cdot3\cdot5\cdots(2n-1)\big]^2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}={}^{4n}C_{2n}:{}^{2n}C_n
\displaystyle =\frac{{}^{4n}C_{2n}}{{}^{2n}C_n}
\displaystyle =\frac{(4n)!}{(2n)!(2n)!}\times\frac{n!n!}{(2n)!}
\displaystyle =\frac{(4n)!(n!)^2}{[(2n)!]^3}
\displaystyle \text{Now,}
\displaystyle (4n)!=\big[1\cdot3\cdot5\cdots(4n-1)\big]\big[2\cdot4\cdot6\cdots4n\big]
\displaystyle =\big[1\cdot3\cdot5\cdots(4n-1)\big]2^{2n}(2n)!
\displaystyle \text{Also,}
\displaystyle (2n)!=\big[1\cdot3\cdot5\cdots(2n-1)\big]\big[2\cdot4\cdot6\cdots2n\big]
\displaystyle =\big[1\cdot3\cdot5\cdots(2n-1)\big]2^n n!
\displaystyle \therefore [(2n)!]^2=\big[1\cdot3\cdot5\cdots(2n-1)\big]^2 2^{2n}(n!)^2
\displaystyle \therefore \text{LHS}=\frac{\big[1\cdot3\cdot5\cdots(4n-1)\big]2^{2n}(2n)!(n!)^2}{(2n)!\big[1\cdot3\cdot5\cdots(2n-1)\big]^2 2^{2n}(n!)^2}
\displaystyle =\frac{1\cdot3\cdot5\cdots(4n-1)}{\big[1\cdot3\cdot5\cdots(2n-1)\big]^2}
\displaystyle =\text{RHS}
\displaystyle \therefore {}^{4n}C_{2n}:{}^{2n}C_n=\big[1\cdot3\cdot5\cdots(4n-1)\big]:\big[1\cdot3\cdot5\cdots(2n-1)\big]^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Evaluate: }{}^{20}C_5+\sum\limits_{r=2}^{5}{}^{25-r}C_4.
\displaystyle \text{Answer:}
\displaystyle {}^{20}C_5+\sum\limits_{r=2}^{5}{}^{25-r}C_4
\displaystyle ={}^{20}C_5+{}^{23}C_4+{}^{22}C_4+{}^{21}C_4+{}^{20}C_4
\displaystyle =\left({}^{20}C_4+{}^{20}C_5\right)+{}^{21}C_4+{}^{22}C_4+{}^{23}C_4
\displaystyle \text{Using }{}^{n}C_{r-1}+{}^{n}C_r={}^{n+1}C_r,
\displaystyle ={}^{21}C_5+{}^{21}C_4+{}^{22}C_4+{}^{23}C_4
\displaystyle ={}^{22}C_5+{}^{22}C_4+{}^{23}C_4
\displaystyle ={}^{23}C_5+{}^{23}C_4
\displaystyle ={}^{24}C_5
\displaystyle =\frac{24!}{5!19!}
\displaystyle =\frac{24\times23\times22\times21\times20}{5\times4\times3\times2\times1}
\displaystyle =4\times23\times22\times21
\displaystyle =42504
\displaystyle \therefore \text{The value of the given expression is }42504.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Let }r\text{ and }n\text{ be positive integers such that }2\leq r\leq n.
\displaystyle \text{Then prove the following:}
\displaystyle \text{i) }\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}
\displaystyle \text{ii) }n\,{}^{n-1}C_{r-1}=(n-r+1)\,{}^{n}C_{r-1}
\displaystyle \text{iii) }\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r}
\displaystyle \text{iv) }{}^{n}C_r+2\,{}^{n}C_{r-1}+{}^{n}C_{r-2}={}^{n+2}C_r
\displaystyle \text{Answer:}
\displaystyle \text{i) LHS}=\frac{{}^{n}C_r}{{}^{n}C_{r-1}}
\displaystyle =\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}
\displaystyle =\frac{n!}{r!(n-r)!}\times\frac{(r-1)!(n-r+1)!}{n!}
\displaystyle =\frac{(r-1)!}{r!}\times\frac{(n-r+1)!}{(n-r)!}
\displaystyle =\frac{1}{r}\times(n-r+1)
\displaystyle =\frac{n-r+1}{r}=\text{RHS}
\displaystyle \therefore \frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}.
\displaystyle \text{ii) LHS}=n\,{}^{n-1}C_{r-1}
\displaystyle =n\left[\frac{(n-1)!}{(r-1)!(n-r)!}\right]
\displaystyle =\frac{n!}{(r-1)!(n-r)!}
\displaystyle \text{RHS}=(n-r+1)\,{}^{n}C_{r-1}
\displaystyle =(n-r+1)\left[\frac{n!}{(r-1)!(n-r+1)!}\right]
\displaystyle =\frac{(n-r+1)n!}{(r-1)!(n-r+1)(n-r)!}
\displaystyle =\frac{n!}{(r-1)!(n-r)!}
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \therefore n\,{}^{n-1}C_{r-1}=(n-r+1)\,{}^{n}C_{r-1}.
\displaystyle \text{iii) LHS}=\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}
\displaystyle =\frac{\frac{n!}{r!(n-r)!}}{\frac{(n-1)!}{(r-1)!(n-r)!}}
\displaystyle =\frac{n!}{r!(n-r)!}\times\frac{(r-1)!(n-r)!}{(n-1)!}
\displaystyle =\frac{n!}{(n-1)!}\times\frac{(r-1)!}{r!}
\displaystyle =n\times\frac{1}{r}
\displaystyle =\frac{n}{r}=\text{RHS}
\displaystyle \therefore \frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r}.
\displaystyle \text{iv) LHS}={}^{n}C_r+2\,{}^{n}C_{r-1}+{}^{n}C_{r-2}
\displaystyle =\left({}^{n}C_r+{}^{n}C_{r-1}\right)+\left({}^{n}C_{r-1}+{}^{n}C_{r-2}\right)
\displaystyle ={}^{n+1}C_r+{}^{n+1}C_{r-1}
\displaystyle ={}^{n+2}C_r=\text{RHS}
\displaystyle \therefore {}^{n}C_r+2\,{}^{n}C_{r-1}+{}^{n}C_{r-2}={}^{n+2}C_r.
\displaystyle \text{Hence proved.}
\displaystyle \\

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