Question 1: Compute:

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

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

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

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

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

$\displaystyle 7! = 7 \times 6!$

$\displaystyle 6! = 6!$

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \\$

Question 3: Find x in each of the following:

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

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

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

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

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

$\displaystyle \Rightarrow x= 36$

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

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

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

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

$\displaystyle \Rightarrow x = 100$

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

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

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

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

$\displaystyle \Rightarrow x = 64$

$\displaystyle \\$

Question 4: Convert the following products into factorials:

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

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

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

$\displaystyle \text{ii) } 3 \cdot 6 \cdot 9 \cdot 12 \cdot 15 \cdot 18 = 3^6 ( 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6) = 3^6 \times 6!$

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

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

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

$\displaystyle \\$

Question 5: Which of the following are true:

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

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

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

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

$\displaystyle \text{Hence } \text{LHS } \neq$ 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.

$\displaystyle \\$

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

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

Hence proved.

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow n = 3$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow n = 9$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow n = 5$

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow 2n = 12$

$\displaystyle \Rightarrow n = 6$

$\displaystyle \\$

Question 11: Prove that:

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{LHS } = \frac{(2n+1)!}{n!}$
$\displaystyle = \frac{(2n+1) (2n)!}{n!}$
$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \ldots n(n+1) \ldots (2n-2) \cdot (2n-1) \cdot (2n) \cdot (2n+1)]}{n!}$
$\displaystyle = \frac{[1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)][2 \cdot 4 \cdot 6 \cdot 8 \ldots (2n-2)(2n)]}{n!}$
$\displaystyle = \frac{[1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)][2^n 1 \cdot 2 \cdot 3 \cdot 4 \ldots (n-1)(n)]}{n!}$
$\displaystyle = 2^n [ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)(2n+1)] = \text{ RHS. Hence proved. }$