Question 1: Compute:

i) \frac{30!}{28!}           ii) \frac{11!-10!}{9!}           iii) LCM ( 6!, 7!, 8!) 

Answer:

i) \frac{30!}{28!} = \frac{30 \times 29 \times 28!}{28!} = 30 \times 29 = 870          

ii) \frac{11!-10!}{9!} = \frac{11 \times 10 \times 9! - 10 \times 9!}{9!} = 11 \times 10 - 10 = 100          

iii) LCM ( 6!, 7!, 8!) 

8! = 8 \times 7 \times 6!

7! = 7 \times 6!

6! = 6!

Therefore LCM ( 6!, 7!, 8!)  is 8 \times 7 \times 6! = 8!

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Question 2: Prove that \frac{1}{9!} + \frac{1}{10!} + \frac{1}{11!} = \frac{122}{11!}

Answer:

LHS = \frac{1}{9!} + \frac{1}{10!} + \frac{1}{11!}

= \frac{1}{9!} + \frac{1}{10 \times 9!} + \frac{1}{11 \times 10 \times 9!}

= \frac{1}{9!} \Big( 1 + \frac{1}{10} + \frac{1}{110} \Big)

= \frac{1}{9!} \Big( \frac{110+11+1}{110}   \Big)

= \frac{122}{11!} = RHS. Hence proved.

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Question 3: Find x in each of the following:

i) \frac{1}{4!} + \frac{1}{5!} = \frac{x}{6!}           ii) \frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}         iii) \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}

Answer:

i)      \frac{1}{4!} + \frac{1}{5!} = \frac{x}{6!}

\Rightarrow \frac{1}{4!} + \frac{1}{5 \times 4!} = \frac{x}{6 \times 5 \times 4!}

\Rightarrow 1 + \frac{1}{5 } = \frac{x}{6 \times 5 }

\Rightarrow \frac{6}{5} = \frac{x}{30}

\Rightarrow x= 36

ii)      \frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}

\Rightarrow \frac{x}{10 \times 9 \times  8!}   = \frac{1}{8!} + \frac{1}{9 \times  8!}

\Rightarrow \frac{x}{90} = 1+ \frac{1}{9}

\Rightarrow \frac{x}{90} = \frac{10}{9}

\Rightarrow x = 100 

iii)     \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}

\Rightarrow \frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8\times 7 \times 6!}

\Rightarrow 1 + \frac{1}{7} = \frac{x}{56}

\Rightarrow \frac{8}{7} = \frac{x}{56}

\Rightarrow x = 64

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Question 4: Convert the following products into factorials:

i) 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10           ii) 3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18

iii) (n+1)(n+2)(n+3) \ldots (2n)           iv) 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)

Answer:

i) 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = \frac{(1 \times 2 \times 3 \times 4 )( 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10)}{1 \times 2 \times 3 \times 4 } = \frac{10!}{4!}          

ii) 3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18 = 3^6 ( 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6)  = 3^6 \times 6!

iii) (n+1)(n+2)(n+3) \ldots (2n) = \frac{(1 \times 2 \times 3 \ldots \times n)[ (n+1)(n+2)(n+3) \ldots (2n) ] }{(1 \times 2 \times 3 \ldots n)} = \frac{(2n)!}{n!}          

iv)     1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)

= \frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1) ]  [2 \cdot 4 \cdot 6 \cdot 8 \ldots 2n  ]  }{ [ 2 \cdot 4 \cdot 6 \cdot 8 \ldots 2n ] } = \frac{(2n)!}{2^n ( 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots n)} = \frac{(2n)!}{2^n n!}

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Question 5: Which of the following are true:

i) (2+3)! = 2! + 3!           ii) (2 \times 3)! = 2! \times 3!

Answer:

i)      (2+3)! = 2! + 3!    

LHS = ( 2 + 3)! = 5! = 120

RHS = 2! + 3! = 2   + 6 = 8

Hence LHS \neq RHS. Hence the statement is false.      

ii)      (2 \times 3)! = 2! \times 3!

LHS = (2 \times 3)! = 6! = 720

RHS = 2! \times 3! = 2 \times 6 = 12

Hence LHS \neq RHS. Hence the statement is false.   

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Question 6: Prove that: n! ( n+2) = n! + ( n+1)!

Answer:

RHS = n! + ( n+1)! = n! + (n+1) n! = n! ( 1+ n+1) = n! ( n+2) = LHS.

Hence proved.

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Question 7: If ( n+2)! = 60[(n-1)!] , find n

Answer:

Given ( n+2)! = 60[(n-1)!]

\Rightarrow (n+2)(n+1)n(n-1)! = 60(n-1)!

\Rightarrow (n+2)(n+1)n = 60

\Rightarrow (n+2)(n+1)n = 5\times 4 \times 3

\Rightarrow n = 3

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Question 8: If ( n+1)! = 90 [( n-1)!] , find n

Answer:

Given ( n+1)! = 90 [( n-1)!]

\Rightarrow (n+1) \times n \times ( n-1)! = 90 ( n-1)!

\Rightarrow n ( n+1) = 90

\Rightarrow n ( n+1) = 9 \times 10

\Rightarrow n = 9

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Question 9: If ( n+3) ! = 56 [(n+1)!] , find n

Answer:

Given ( n+3) ! = 56 [(n+1)!]

\Rightarrow (n+3)(n+2)(n+1)! = 56 (n+1)!

\Rightarrow (n+3)(n+2) = 56

\Rightarrow (n+3)(n+2) = 8\times 7

\Rightarrow n = 5

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Question 10: If \frac{(2n)!}{3! ( 2n-3)!} and \frac{n!}{2! ( n-2)!} are in the ratio 44:3 , find n

Answer: 

\frac{(2n)!}{3! ( 2n-3)!}  : \frac{n!}{2! ( n-2)!}  = 44:3

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

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

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

\Rightarrow \frac{4(2n-1)}{3} \times \frac{2!}{n(n-1)}  = \frac{44}{3} 

\Rightarrow 2n-1 = 11

\Rightarrow 2n = 12

\Rightarrow n = 6

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Question 11: Prove that:

i) \frac{n!}{(n-r)!} = n(n-1)(n-2) \ldots (n-(n-r))

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

Answer:

i)       \frac{n!}{(n-r)!} = n(n-1)(n-2) \ldots (n-(n-r))

LHS = \frac{n!}{(n-r)!} = \frac{n(n-1)(n-2) \ldots [(n - ( r-1)](n-r)!}{(n-r)!}

= n(n-1)(n-2) \ldots [(n - ( r-1)] = RHS. Hence proved.

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

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

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

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

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

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

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

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

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Question 12: Prove that: \frac{(2n+1)!}{n!} = 2^n \Big\{ 1\cdot 3 \cdot 5 \ldots (2n-1)(2n+1)  \Big\}

Answer:

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

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

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

= \frac{[1 \cdot 3 \cdot 5 \cdot 7  \ldots (2n-1)(2n+1)][2 \cdot 4 \cdot 6 \cdot 8 \ldots (2n-2)(2n)]}{n!}

= \frac{[1 \cdot 3 \cdot 5 \cdot 7  \ldots (2n-1)(2n+1)][2^n 1 \cdot 2 \cdot 3 \cdot 4  \ldots (n-1)(n)]}{n!}

= 2^n [ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)] = RHS. Hence proved.

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