Question 1: Compute:

\displaystyle \text{i) } \frac{30!}{28!}  \hspace{1.0cm}  \text{ii) } \frac{11!-10!}{9!}  \hspace{1.0cm}  \text{iii) } \text{LCM } ( 6!, 7!, 8!)

Answer:

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

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

\displaystyle \text{iii) } \text{LCM } ( 6!, 7!, 8!)

\displaystyle 8! = 8 \times 7 \times 6!

\displaystyle 7! = 7 \times 6!

\displaystyle 6! = 6!

\displaystyle \text{Therefore } \text{LCM } ( 6!, 7!, 8!) \text{ is }  8 \times 7 \times 6! = 8!

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

Answer:

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

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

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

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

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

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Question 3: Find x in each of the following:

\displaystyle \text{i) } \frac{1}{4!} + \frac{1}{5!} = \frac{x}{6!}  \hspace{1.0cm}  \text{ii) } \frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}  \hspace{1.0cm}  \text{iii) } \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}  

Answer:

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

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

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

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

\displaystyle \Rightarrow x= 36

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

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

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

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

\displaystyle \Rightarrow x = 100

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

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

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

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

\displaystyle \Rightarrow x = 64

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Question 4: Convert the following products into factorials:

\displaystyle \text{i) } 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10  \hspace{1.0cm}  \text{ii) } 3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18

\displaystyle \text{iii) } (n+1)(n+2)(n+3) \ldots (2n)  \hspace{1.0cm}  \text{iv) } 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)

Answer:

\displaystyle \text{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!}  

\displaystyle \text{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!

\displaystyle \text{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!}  

\displaystyle \text{iv) } 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \ldots (2n-1)

\displaystyle = \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:

\displaystyle \text{i) } (2+3)! = 2! + 3!  \hspace{1.0cm}  \text{ii) } (2 \times 3)! = 2! \times 3!

Answer:

\displaystyle \text{i) } (2+3)! = 2! + 3!  

\displaystyle \text{LHS } = ( 2 + 3)! = 5! = 120

\displaystyle \text{RHS } = 2! + 3! = 2 + 6 = 8

\displaystyle \text{Hence } \text{LHS } \neq RHS. Hence the statement is false. 

\displaystyle \text{ii) } (2 \times 3)! = 2! \times 3!

\displaystyle \text{LHS } = (2 \times 3)! = 6! = 720

\displaystyle \text{RHS } = 2! \times 3! = 2 \times 6 = 12

\displaystyle \text{Hence } \text{LHS } \neq RHS. Hence the statement is false. 

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\displaystyle \text{Question 6: Prove that: } n! ( n+2) = n! + ( n+1)!

Answer:

\displaystyle \text{RHS } = n! + ( n+1)! = n! + (n+1) n! = n! ( 1+ n+1) = n! ( n+2) = \text{ LHS. }

Hence proved.

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\displaystyle \text{Question 7: If } ( n+2)! = 60[(n-1)!] , \text{ find } n

Answer:

\displaystyle \text{Given } ( n+2)! = 60[(n-1)!]

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

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

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

\displaystyle \Rightarrow n = 3

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\displaystyle \text{Question 8: If } ( n+1)! = 90 [( n-1)!] , \text{ find } n

Answer:

\displaystyle \text{Given } ( n+1)! = 90 [( n-1)!]

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

\displaystyle \Rightarrow n ( n+1) = 90

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

\displaystyle \Rightarrow n = 9

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\displaystyle \text{Question 9: If } ( n+3) ! = 56 [(n+1)!] , \text{ find } n

Answer:

\displaystyle \text{Given } ( n+3) ! = 56 [(n+1)!]

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

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

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

\displaystyle \Rightarrow n = 5

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

Answer:

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

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

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

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

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

\displaystyle \Rightarrow 2n-1 = 11

\displaystyle \Rightarrow 2n = 12

\displaystyle \Rightarrow n = 6

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Question 11: Prove that:

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

\displaystyle = \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!}  

\displaystyle = \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!}  

\displaystyle = \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!}  

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