Question 1: Compute:

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

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 $\frac{1}{9!}$ $+$ $\frac{1}{10!}$ $+$ $\frac{1}{11!}$ $=$ $\frac{122}{11!}$

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

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

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) $(2 \times 3)! = 2! \times 3!$

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: $n! ( n+2) = n! + ( n+1)!$

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 $n$

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 $n$

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 $n$

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 $n$

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

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

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

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

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

$\Rightarrow 2n-1 = 11$

$\Rightarrow 2n = 12$

$\Rightarrow n = 6$

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Question 11: Prove that:

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

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

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

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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