\displaystyle \textbf{Question 1: } \text{Compute:}
\displaystyle \text{i) }\frac{30!}{28!}\qquad \text{ii) }\frac{11!-10!}{9!}\qquad \text{iii) }\text{LCM}(6!,7!,8!)
\displaystyle \text{Answer:}
\displaystyle \text{i) }\frac{30!}{28!}=\frac{30\times29\times28!}{28!}=30\times29=870
\displaystyle \\

\displaystyle \text{ii) }\frac{11!-10!}{9!}=\frac{10!(11-1)}{9!}=\frac{10!\times10}{9!}=10\times10=100
\displaystyle \\

\displaystyle \text{iii) }\text{Since }6!\mid7!\mid8!,\text{ the LCM}(6!,7!,8!)=8!.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Prove that }\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}=\frac{122}{11!}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}
\displaystyle =\frac{1}{9!}+\frac{1}{10\times9!}+\frac{1}{11\times10\times9!}
\displaystyle =\frac{1}{9!}\left(1+\frac{1}{10}+\frac{1}{110}\right)
\displaystyle =\frac{1}{9!}\left(\frac{110+11+1}{110}\right)
\displaystyle =\frac{122}{110\times9!}=\frac{122}{11!}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Find }x\text{ in each of the following:}
\displaystyle \text{i) }\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!}\qquad \text{ii) }\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!}\qquad \text{iii) }\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}
\displaystyle \text{Answer:}
\displaystyle \text{i) }\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!}
\displaystyle \Rightarrow \frac{1}{4!}+\frac{1}{5\times4!}=\frac{x}{6\times5\times4!}
\displaystyle \Rightarrow 1+\frac{1}{5}=\frac{x}{30}
\displaystyle \Rightarrow \frac{6}{5}=\frac{x}{30}
\displaystyle \Rightarrow x=36
\displaystyle \\

\displaystyle \text{ii) }\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!}
\displaystyle \Rightarrow \frac{x}{10\times9\times8!}=\frac{1}{8!}+\frac{1}{9\times8!}
\displaystyle \Rightarrow \frac{x}{90}=1+\frac{1}{9}
\displaystyle \Rightarrow \frac{x}{90}=\frac{10}{9}
\displaystyle \Rightarrow x=100
\displaystyle \\

\displaystyle \text{iii) }\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}
\displaystyle \Rightarrow \frac{1}{6!}+\frac{1}{7\times6!}=\frac{x}{8\times7\times6!}
\displaystyle \Rightarrow 1+\frac{1}{7}=\frac{x}{56}
\displaystyle \Rightarrow \frac{8}{7}=\frac{x}{56}
\displaystyle \Rightarrow x=64
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Convert the following products into factorials:}
\displaystyle \text{i) }5\cdot6\cdot7\cdot8\cdot9\cdot10\qquad \text{ii) }3\cdot6\cdot9\cdot12\cdot15\cdot18
\displaystyle \text{iii) }(n+1)(n+2)(n+3)\ldots(2n)\qquad \text{iv) }1\cdot3\cdot5\cdot7\cdot9\ldots(2n-1)
\displaystyle \text{Answer:}
\displaystyle \text{i) }5\cdot6\cdot7\cdot8\cdot9\cdot10=\frac{(1\cdot2\cdot3\cdot4)(5\cdot6\cdot7\cdot8\cdot9\cdot10)}{1\cdot2\cdot3\cdot4}=\frac{10!}{4!}
\displaystyle \\

\displaystyle \text{ii) }3\cdot6\cdot9\cdot12\cdot15\cdot18=3^6(1\cdot2\cdot3\cdot4\cdot5\cdot6)=3^6\times6!
\displaystyle \\

\displaystyle \text{iii) }(n+1)(n+2)(n+3)\ldots(2n)
\displaystyle =\frac{(1\cdot2\cdot3\ldots n)\left[(n+1)(n+2)\ldots(2n)\right]}{1\cdot2\cdot3\ldots n}=\frac{(2n)!}{n!}
\displaystyle \\

\displaystyle \text{iv) }1\cdot3\cdot5\cdot7\cdot9\ldots(2n-1)
\displaystyle =\frac{\left[1\cdot3\cdot5\cdot7\ldots(2n-1)\right]\left[2\cdot4\cdot6\cdot8\ldots2n\right]}{2\cdot4\cdot6\cdot8\ldots2n}
\displaystyle =\frac{(2n)!}{2^n(1\cdot2\cdot3\ldots n)}=\frac{(2n)!}{2^nn!}
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Which of the following are true:}
\displaystyle \text{i) }(2+3)!=2!+3!\qquad \text{ii) }(2\times3)!=2!\times3!
\displaystyle \text{Answer:}
\displaystyle \text{i) }(2+3)!=2!+3!
\displaystyle \text{LHS}=(2+3)!=5!=120
\displaystyle \text{RHS}=2!+3!=2+6=8
\displaystyle \therefore \text{LHS}\ne\text{RHS}.\text{ Hence the statement is false.}
\displaystyle \\

\displaystyle \text{ii) }(2\times3)!=2!\times3!
\displaystyle \text{LHS}=(2\times3)!=6!=720
\displaystyle \text{RHS}=2!\times3!=2\times6=12
\displaystyle \therefore \text{LHS}\ne\text{RHS}.\text{ Hence the statement is false.}
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Prove that }n!(n+2)=n!+(n+1)!.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=n!(n+2)
\displaystyle =n!(n+1+1)
\displaystyle =(n+1)n!+n!
\displaystyle =(n+1)!+n!=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{If }(n+2)!=60[(n-1)!],\text{ find }n.
\displaystyle \text{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\times4\times3
\displaystyle \Rightarrow n=3
\displaystyle \text{(Verification: }3\times4\times5=60\text{)}
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{If }(n+1)!=90[(n-1)!],\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \text{Given }(n+1)!=90(n-1)!
\displaystyle \Rightarrow (n+1)n(n-1)!=90(n-1)!
\displaystyle \Rightarrow n(n+1)=90
\displaystyle \Rightarrow n(n+1)=9\times10
\displaystyle \Rightarrow n=9
\displaystyle \text{(Verification: }9\times10=90\text{)}
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{If }(n+3)!=56[(n+1)!],\text{ find }n.
\displaystyle \text{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\times7
\displaystyle \Rightarrow n+3=8\text{ and }n+2=7
\displaystyle \Rightarrow n=5
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{If }\frac{(2n)!}{3!(2n-3)!}\text{ and }\frac{n!}{2!(n-2)!}\text{ are in the}
\displaystyle \text{ratio }44:3,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \frac{(2n)!}{3!(2n-3)!}:\frac{n!}{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{(2n)(2n-1)\cdot2(n-1)}{6}\times\frac{2}{n(n-1)}=\frac{44}{3}
\displaystyle \Rightarrow \frac{4(2n-1)}{3}=\frac{44}{3}
\displaystyle \Rightarrow 4(2n-1)=44
\displaystyle \Rightarrow 2n-1=11
\displaystyle \Rightarrow 2n=12
\displaystyle \Rightarrow n=6
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Prove that:}
\displaystyle \text{i) }\frac{n!}{(n-r)!}=n(n-1)(n-2)\ldots(n-r+1)
\displaystyle \text{ii) }\frac{n!}{(n-r)!r!}+\frac{n!}{(n-r+1)!(r-1)!}=\frac{(n+1)!}{(n-r+1)!r!}
\displaystyle \text{Answer:}
\displaystyle \text{i) LHS}=\frac{n!}{(n-r)!}
\displaystyle =\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}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \text{ii) 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)!}\left[\frac{1}{r}+\frac{1}{n-r+1}\right]
\displaystyle =\frac{n!}{(n-r)!(r-1)!}\left[\frac{n-r+1+r}{r(n-r+1)}\right]
\displaystyle =\frac{n!}{(n-r)!(r-1)!}\left[\frac{n+1}{r(n-r+1)}\right]
\displaystyle =\frac{(n+1)!}{r!(n-r+1)(n-r)!}
\displaystyle =\frac{(n+1)!}{(n-r+1)!r!}=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Prove that }\frac{(2n+1)!}{n!}=2^n\left\{1\cdot3\cdot5\ldots(2n-1)(2n+1)\right\}.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(2n+1)!}{n!}
\displaystyle =\frac{1\cdot2\cdot3\cdot4\ldots(2n)(2n+1)}{n!}
\displaystyle =\frac{\left[1\cdot3\cdot5\ldots(2n-1)(2n+1)\right]\left[2\cdot4\cdot6\ldots(2n)\right]}{n!}
\displaystyle =\frac{\left[1\cdot3\cdot5\ldots(2n-1)(2n+1)\right]\left[2^n(1\cdot2\cdot3\ldots n)\right]}{n!}
\displaystyle =\frac{2^nn!\left[1\cdot3\cdot5\ldots(2n-1)(2n+1)\right]}{n!}
\displaystyle =2^n\left[1\cdot3\cdot5\ldots(2n-1)(2n+1)\right]=\text{RHS}.
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

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