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}

Answer:

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 .

Answer:

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} .

Answer:

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} .

Answer:

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 .

Answer:

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 .

Answer:

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 .

Answer:

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 .

Answer:

^{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 .

Answer:

^{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 .

Answer:

^{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 .

Answer:

^{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 .

Answer:

^{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 .

Answer:

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} .

Answer:

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}

Answer:

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)!

Answer:

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) 

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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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