Question 1: Evaluate the following:

i) $^{14} \rm C_3$     ii) $^{12} \rm C_{10}$     iii) $^{35} \rm C_{35}$     iv) $^{n+1} \rm C_n$    v) $\sum \limits_{r=1}^{5}$  $^{5} \rm C_{r}$

i) $^{14} \rm C_3 =$ $\frac{14!}{3! (14-3)!}$ $=$ $\frac{14!}{3! 11!}$ $=$ $\frac{14 \times 13 \times 12}{6}$ $= 364$

ii) $^{12} \rm C_{10} =$ $\frac{12!}{10! (12-10)!}$ $=$ $\frac{12!}{10! 2!}$ $=$ $\frac{11 \times 11 }{2}$ $= 66$

iii) $^{35} \rm C_{35} =$ $\frac{35!}{35! (35-35)!}$ $=$ $\frac{35!}{35! 0!}$ $= 1$

iv) $^{n+1} \rm C_n =$ $\frac{(n+1)!}{n! (n+1-n)!}$ $=$ $\frac{(n+1)n!}{n!}$ $= (n+1)$

v) $\sum \limits_{r=1}^{5}$  $^{5} \rm C_{r}$ $= ^{5} \rm C_{1} + ^{5} \rm C_{2} + ^{5} \rm C_{3}+^{5} \rm C_{4}+^{5} \rm C_{5}$

$=$ $\frac{5!}{5! (5-1)!}$ $+$ $\frac{5!}{5! (5-2)!}$ $+$ $\frac{5!}{5! (5-3)!}$ $+$ $\frac{5!}{5! (5-4)!}$ $+$ $\frac{5!}{5! (5-5)!}$

$= 5 + 10 + 10 + 5 + 1 = 31$

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Question 2: If $^{n} \rm C_{12} = ^{n} \rm C_{5}$, find the value of $n$.

Given $^{n} \rm C_{12} = ^{n} \rm C_{5}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $12 + 5 = n \Rightarrow n = 17$

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Question 3: If $^{n} \rm C_{4} = ^{n} \rm C_{6}$, find $^{12} \rm C_{n}$.

Given $^{n} \rm C_{4} = ^{n} \rm C_{6}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $4 + 6 = n \Rightarrow n = 10$

$^{12} \rm C_{10} =$ $\frac{12!}{10! 2!}$ $=$ $\frac{12 \times 11}{2}$ $= 66$

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Question 4: If $^{n} \rm C_{10} = ^{n} \rm C_{12}$, find $^{23} \rm C_{n}$.

Given $^{n} \rm C_{10} = ^{n} \rm C_{12}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $10 + 12 = n \Rightarrow n = 22$

$^{23} \rm C_{22} =$ $\frac{23!}{22! (23-22)!}$ $=$ $\frac{23! }{22!}$ $= 23$

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Question 5: If $^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$, find $x$.

Given $^{24} \rm C_{x} = ^{24} \rm C_{2x+3}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $x + 2x+3 = 24 \Rightarrow x = 7$

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Question 6: If $^{18} \rm C_{x} = ^{18} \rm C_{x+2}$, find $x$.

Given $^{18} \rm C_{x} = ^{18} \rm C_{x+2}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $x + x+2 = 18 \Rightarrow x = 8$

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Question 7: If $^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$, find $r$.

Given $^{15} \rm C_{3r} = ^{15} \rm C_{r+3}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $3r+r+3 = 15 \Rightarrow r = 3$

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Question 8: If $^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$, find $r$.

$^{8} \rm C_{r} - ^{7} \rm C_{3} = ^{7} \rm C_{2}$

$\Rightarrow ^{8} \rm C_{r} = ^{7} \rm C_{2} + ^{7} \rm C_{3}$

$\Rightarrow$ $\frac{8!}{r! (8-r)!}$ $=$ $\frac{7!}{2! 5!}$ $+$ $\frac{7!}{3! 4!}$

$\Rightarrow$ $\frac{8!}{r! (8-r)!}$ $=$ $\frac{7 \times 6}{2}$ $+$ $\frac{7 \times 6 \times 5}{6}$

$\Rightarrow$ $\frac{8!}{r! (8-r)!}$ $= 56$

$\Rightarrow r! (8-r)! = 720$

$\Rightarrow r! (8-r)! = 3! \times 5!$

Comparing LHS with RHS we get,

If $r = 3$  and $8-r = 5 \Rightarrow r = 3$

If $r = 5$, then $8-r = 3 \Rightarrow r = 5$

Hence $r$ can be $3$ or $5$.

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Question 9: If $^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$, find $r$.

$^{15} \rm C_{r} : ^{15} \rm C_{r-1} = 11:5$

$\Rightarrow$ $\frac{15!}{r! (15-r)!}$ $\times$ $\frac{(r-1)! ( 15-r+1)!}{15!}$ $=$ $\frac{11}{5}$

$\Rightarrow$ $\frac{(r-1)! ( 16-r) (15-r)!}{r(r-1)!(15-r)!}$ $=$ $\frac{11}{5}$

$\Rightarrow$ $\frac{16-r}{r}$ $=$ $\frac{11}{5}$

$\Rightarrow 80-5r = 11r$

$\Rightarrow 80 = 16 r$

$\Rightarrow r = 5$

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Question 10: If $^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$, find $r$.

$^{n+2} \rm C_{8} : ^{n-2} \rm P_{4} = 57:16$

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

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

$\Rightarrow (n+2)(n+1)(n)(n-1) =$ $\frac{8!}{16}$ $\times 57 = 7 \times 6 \times 5 \times 4 \times 3 \times 19 \times 3$

$= 18 \times 19 \times 20 \times 21$

$\Rightarrow n = 18$

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Question 11: If $^{28} \rm C_{2r} : ^{24} \rm C_{2r-4} = 225:11$, find $r$.

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

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

$\Rightarrow$ $\frac{28 \times 27 \times 26 \times 25 \times ( 2r-4)!}{(2r) ( 2r-1)(2r-2)(2r-3)(2r-4)!}$ $=$ $\frac{225}{11}$

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

$\Rightarrow (2r) ( 2r-1)(2r-2)(2r-3) = 28 \times 33 \times 26 = 11 \times 12 \times 13 \times 14$

$\Rightarrow 2r-3 = 11 \Rightarrow r = 7$

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Question 12: If  $^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $^{n} \rm C_{6}$ are in AP, then find $n$.

$^{n} \rm C_{4}, \ \ ^{n} \rm C_{5}$ and $^{n} \rm C_{6}$ are in AP

$\Rightarrow ^{n} \rm C_{5} - ^{n} \rm C_{4} = ^{n} \rm C_{6} - ^{n} \rm C_{5}$

$\Rightarrow$ $\frac{n!}{5! (n-5)!}$ $-$ $\frac{n!}{4! (n-4)!}$ $=$ $\frac{n!}{6! (n-6)!}$ $-$ $\frac{n!}{5! (n-5)!}$

$\Rightarrow$ $\frac{1}{5! (n-5)!}$ $-$ $\frac{1}{4! (n-4)!}$ $=$ $\frac{1}{6! (n-6)!}$ $-$ $\frac{1}{5! (n-5)!}$

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

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

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

$\Rightarrow$ $\frac{1}{n-5}$ $\Big[$ $\frac{2}{5}$ $-$ $\frac{1}{n-4}$ $\Big] =$ $\frac{1}{30}$

$\Rightarrow$ $\frac{1}{n-5}$ $\Big[$ $\frac{2n-8-5}{5(n-4)}$ $\Big] =$ $\frac{1}{30}$

$30(2n-13) = 5 ( n-4)(n-5)$

$12n - 78 = n^2 - 9n + 20$

$n^2 - 21n +98 = 0$

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

$n = 7$ or $n = 14$

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Question 13: If $^{2n} \rm C_{3} : ^{n} \rm C_{2} = 44:3$, find $n$.

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

$\Rightarrow$ $\frac{(2n)!}{3! (2n-3)!}$ $\times$ $\frac{2! (n-2)!}{n!}$ $=$ $\frac{44}{3}$

$\Rightarrow$ $\frac{(2n)(2n-1)(2n-2)}{3!}$ $\times$ $\frac{2!}{n(n-1)}$ $=$ $\frac{44}{3}$

$\Rightarrow$ $\frac{(2n-1)(2n-2)}{2(n-1)}$ $= 11$

$2n-1 = 11 \Rightarrow n = 6$

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Question 14: If $^{16} \rm C_{r} = ^{16} \rm C_{r+2}$, find $^{r} \rm C_{4}$.

Given $^{16} \rm C_{r} = ^{16} \rm C_{r+2}$

Applying the formula, when $^{n} \rm C_{p} = ^{n} \rm C_{q}$ $\Rightarrow p+q = n$

Therefore $r + ( r + 2) = 16 \Rightarrow r = 7$

$^{7} \rm C_{4} =$ $\frac{7!}{4! (7-4)!}$ $=$ $\frac{7 \times 6 \times 5 }{6}$ $= 35$

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Question 15: If $\alpha = ^{m} \rm C_{2}$, then find the value of $^{\alpha} \rm C_{2}$

Given $\alpha = ^{m} \rm C_{2} \Rightarrow \alpha =$ $\frac{m!}{2! (m-2)!}$ $=$ $\frac{m(m-1)}{2}$

$^{\alpha} \rm C_{2} =$ $\frac{\alpha !}{2! (\alpha - 2)!}$

$=$ $\frac{\alpha ( \alpha - 1)}{2}$

$=$ $\frac{1}{2}$ $\times \frac{m(m-1)}{2}$ $\times \Big[$ $\frac{m(m-1)}{2}$ $- 1 \Big]$

$=$ $\frac{m(m-1)}{4}$ $\times \Big[$ $\frac{m^2 - m - 2}{2}$ $\Big]$

$=$ $\frac{1}{8}$ $[ (m-1)(m^3 - m^2 - 2m) ]$

$=$ $\frac{1}{8}$ $[ m^4 - m^3 - 2m^2 - m^3 + m^2+ 2m]$

$=$ $\frac{1}{2}$ $[ m^4 - 2m^3 - m^2 + 2m ]$

$=$ $\frac{1}{8}$ $(m^2-2m)(m^2 -1)$

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

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Question 16: Prove that the product of $2n$ consecutive negative integers is divisible by $( 2n)!$

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

Product of $2n$ negative integers $= ( -1)^{2n} [ r ( r+1) (r+2) \ldots ( r+2n-1)]$

$= (-1)^{2n} \Big[$ $\frac{(r-1)! r ( r+1) (r+2) \ldots ( r+2n-1)}{(r-1)!}$ $\Big]$

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

$= ^{r+2n-1} \rm C_{2n} \cdot (2n)!$

Therefore product of $2n$ consecutive negative integers is divisible by $( 2n)!$

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Question 17: For all positive integers $n$, show that

$^{2n} \rm C_n + ^{2n} \rm C_{n-1} =$ $\frac{1}{2}$ $\big( ^{2n+2} \rm C_{n+1} \big)$

LHS $= ^{2n} \rm C_n + ^{2n} \rm C_{n-1}$

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

$=$ $\frac{(2n)!}{(n-1)! n!}$ $\Big[$ $\frac{1}{n}$ $+$ $\frac{1}{n+1}$ $\Big]$

$=$ $\frac{(2n)!}{(n-1)! n!}$ $\Big[$ $\frac{2n+1}{n(n+1)}$ $\Big]$

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

RHS $=$ $\frac{1}{2}$ $\Big[ ^{2n+2} \rm C_{n+1} \Big]$

$=$ $\frac{1}{2}$ $\Big[$ $\frac{(2n+2)!}{(n+1)! (n+1)!}$ $\Big]$

$=$ $\frac{1}{2}$ $\Big[$ $\frac{(2n+2)(2n+1)!}{(n+1) n! (n+1)!}$ $\Big]$

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

Therefore LHS = RHS hence proved.

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Question 18: Prove that: $^{4n} \rm C_{2n} : ^{2n} \rm C_{n} = [ 1 \cdot 3 \cdot 5 \ldots (4n-1) ] : [ 1 \cdot 3 \cdot 5 \ldots (2n-1) ]^2$

LHS $= ^{4n} \rm C_{2n} : ^{2n} \rm C_{n}$

$=$ $\frac{^{4n} \rm C_{2n}}{^{2n} \rm C_{n}}$

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

$=$ $\frac{(4n)(4n-1)(4n-2) \ldots (3) (2) (1)}{[ (2n)(2n-1)(2n-2) \ldots (3)(2)(1)]^2}$ $\frac{(n!)^2}{(2n)!}$

$=$ $\frac{(n!)^2}{(2n)!}$ $\frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] [ 2 \cdot 4 \cdot 6 \ldots (4n-2)(4n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 [ 2 \cdot 4 \cdot 6 \ldots (2n-2)(2n) ]^2 }$

$=$ $\frac{(n!)^2}{(2n)!}$ $\frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (2n-1)(2n) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 2^{2n} [ 1 \cdot 2 \cdot 3 \ldots (n-1)(n) ]^2 }$

$=$ $\frac{(n!)^2}{(2n)!}$ $\frac{(2n)!}{(n!)^2}$ $\frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 }$

$=$ $\frac{ [ 1 \cdot 3 \cdot 5 \ldots (4n-3)(4n-1) ] }{ [ 1 \cdot 3 \cdot 5 \ldots (2n-3)(2n-1) ]^2 }$ $=$ RHS. Hence proved.

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Question 19: Evaluate: $^{20} \rm C_{5} + \sum \limits_{r=2}^{5}$ $^{25-r} \rm C_{4}$

LHS $= ^{20} \rm C_{5} + \sum \limits_{r=2}^{5}$

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

$= \big( ^{20} \rm C_{4} + ^{20} \rm C_{5} \big) + \big( ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4} \big)$

Since $^{n} \rm C_{r-1} + ^{n} \rm C_{r} = ^{n+1} \rm C_{r}$

$= ^{21} \rm C_{5} + ^{21} \rm C_{4} +^{22} \rm C_{4} +^{23} \rm C_{4}$

$= ^{22} \rm C_{5} + ^{22} \rm C_{4} +^{23} \rm C_{4}$

$= ^{23} \rm C_{5} + ^{23} \rm C_{4}$

$= ^{24} \rm C_{5}$

$=$ $\frac{24!}{5! 9!}$ $=$ $\frac{24 \times 23 \times 22 \times 21 \times 20}{5 \times 4 \times 3 \times 2 \times 1 }$ $= 4 \times 23 \times 22 \times 21 = 41504$

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Question 20: Let $r$ and $n$ be positive integers such that $1 \leq r \leq n$. The prove the following:

i) $\frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}}$ $=$ $\frac{n-r+1}{r}$

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

iii) $\frac{^{n} \rm C_{r}}{^{n-1} \rm C_{r-1}}$ $=$ $\frac{n}{r}$

iv) $^{n} \rm C_{r} + 2 ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2} = ^{n+2} \rm C_{r}$

i) $\frac{^{n} \rm C_{r}}{^{n} \rm C_{r-1}}$ $=$ $\frac{n!}{r! (n-r)!}$ $\times$ $\frac{(r-1)! (n-r+1)!}{n!}$ $=$ $\frac{(r-1)!}{r! (r-1)!}$ $\frac{(n-r+1)! (n-r)!}{(n-r)!}$ $=$ $\frac{n-r+1}{r}$

ii) LHS $= n \cdot ^{n-1} \rm C_{r-1} =$ $\frac{n (n-1)!}{(r-1)! (n-1-r+1)!}$ $=$ $\frac{n!}{(r-1)!(n-r)!}$

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

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

Therefore LHS $=$ RHS

iii) LHS $=$ $\frac{^{n} \rm C_{r}}{^{n-1} \rm C_{r-1}}$

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

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

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

$=$ $\frac{n}{r}$ $=$ RHS

iv) LHS $= ^{n} \rm C_{r} + 2 ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2}$

$= \big( ^{n} \rm C_{r} + ^{n} \rm C_{r-1} \big) + \big( ^{n} \rm C_{r-1} + ^{n} \rm C_{r-2} \big)$

$= ^{n+1} \rm C_{r} + ^{n+1} \rm C_{r-1}$

$= ^{n+2} \rm C_{r} =$ RHS. Hence proved.

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