Tutorial:
Basic Permutation & Combination

Question

Tutorial:
Basic Permutation & Combination

rvc · Answer

$\Large\color{red}{\bf \diamondsuit What~is~the~difference~between~Permutation}$

$\Large\color{red}{\bf \&~Combination?}$

$\large\color{black}{\rm The~only~difference~is~the~ORDER.~In~permutation~order~is}$

$\large\color{black}{\rm Important.~The~word~combination~itself~tells~us~that~the~order~}$

$\large\color{black}{\rm doesn't~matter.Permutation~is~the~arrangement~of~objects~in}$

$\large\color{black}{\rm definite~order.}$

rvc · Answer

$\Large\color{purple}{\sf To~remember~:~\color{goldenrod}Permutation~...\color{goldenrod}Position}$

           $\bbox [20px, goldenrod, border:5pt solid black]{\Large\color{white}{ \rm A~Permutation~is~an~{\bf \color{yellow}{ordered}}~ Combination.}}$

rvc · Answer

$\Large\color{purple}{\sf \diamondsuit~ If~a~task~can~be~done~in~\color{black}{m}~ways~\&~another~in~\color{black}{n}~ways~}$
$\Large\color{purple}{\sf then~\color{black}{both~Together~can~be~done~in}~\bbox[5pt,goldenrod, border:2pt solid black ]{\huge m~x ~n}~ways.}$

rvc · Answer

$\Large\color{black}{\rm Example:}$

$\Large\color{red}{\bf \diamondsuit~ David~has~3~shirts~\&~2~sweaters.In~how\~many~ways~he~can~wear~1~shirt~and~1~sweater? }$

$\large\color{black}{\rm David~can~wear~one~shirt~with~one~of~the~2~sweaters.So~with~1~\shirt~he~can~wear~the~sweater~in~2~ways.Now~there~are~3~shirts~so~\he~can~wear~them~in~6~ways.\\hspace{250px}OR\~Here~let~m~be~the~number~of~shirts~and~n~be~the~number~of~sweaters.\So~both~the~tasks~together~can~be~done~in~\color{purple}{m~ x~n}~ways.So~m=3\and~n=2~\therefore  Total~ways~:3~x~2 =6~ways }$

rvc · Answer

$\Large\color{purple}{\sf \diamondsuit~ If~a~task~can~be~done~in~\color{black}{m}~ways~\&~another~in~\color{black}{n}~ways~\~then~exactly~one~of~the~events~can~be~done\~\color{black}{(m+n)}~ways}$

rvc · Answer

$\Large\color{black}{\rm Example:}$

$\Large\color{red}{\bf \diamondsuit~ David~has~3~shirts~\&~2~sweaters.In~how\~many~ways~he~can~wear~either~a~shirt~or~a\~sweater? }$

$\large\color{black}{\rm David~can~wear~anyone~of~the~shirt~so~there~are~3~shirts~\so~there~are~3~ways.~Similarily~2~sweaters~can~be~worn~in~2~ways.\Since~he~is~asked~to~wear~one~at~a~time~so~in~total~there~are~5~ways.}$
$\hspace{250px}{\large\color{black}{\rm OR}}$
$\large\color{black}{\rm Here~let~m:no~of~shirts~\&~n:no~of~sweaters.~Since~the~task~is~to~be\~performed~one~at~a~time~so~total~no~of~ways~is~\color{purple}{(m+n)}~ways.\~\therefore~Total~no~of~ways:3+2=5~ways.}$

anonymous · Answer

Here's an example to more clearly demonstrate the difference between a permutation and combination.

Say you're making arrangements of 3 objects taken 3 at a time. Label these $A,B,C$.

Here is a list of all the possible arrangements. Arrangements = permutations if each are distinct from one another.
$$\begin{matrix}
\color{red}{ABC}&BAC&CAB\
ACB&BCA&CBA
\end{matrix}$$
We have 6 total permutations, but only 1 combination. A combination is a string of distinct letters, whereas permutations don't exclude repeats.

Indeed,$${}_3P{}_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=3!=6$$and$${}_3C{}_3=\frac{3!}{3!(3-3)!}=\frac{3!}{3!0!}=1$$
Notice the relationship between the number of permutations and combinations:
$${}_{n}P{}_{k}=k!{}_{n}C{}_{k}$$
Of the $k$ letters, you have $k!$ ways of arranging them in a string, so multiplying this by the number of combinations will give you the number of ways the distinct letters can be arranged relative to each other.

rvc · Answer

$\bbox[10pt, black, border:2pt solid red]{\color{white} {\bf\huge  \bigstar~~ \huge \underline {Permutation}~~\bigstar }}$

rvc · Answer

$\Large\color{purple}{\rm There~are~basically~two~types~of~permutation~:\ \circ \color{blue}{\bf ~Repetition~is~Allowed:}\color{orange}{\rm such~as~number~lock.\ \hspace{269px}{~It~could be "333".}  }\ \circ \color{blue}{\bf No~Repetition:}\color{orange}{\rm for~example~the~first~three~\ \hspace{190px}{people~in~a~running~race.~You~\ \hspace{190px}{can't~be~first~ and~ second.}}}}$

rvc · Answer

$\Large\color{red}{\bf \diamondsuit~ 1.~ Permutations~with~ Repetition~\diamondsuit  }$

$\large\color{black}{\rm Suppose~we~have~\color{blue}n~objects~to~choose~from.......then~we~have~\\color{blue}n~choices~each~time.\~ \When~choosing~\color{brown}r~of~them, the~permutations~are: \ ~\~\hspace{120pt} \bf n~x~n~x~...(r~times)}$

rvc · Answer

$\large\color{black}{\rm In~other~words, there~are~\color{blue}n~possibilities~for~the~\color{gold}{first~choice},~\THEN~there~are~\color{blue}n~ possibilites~for~the~\color{gold}{second~choice}, and~so~on, \multplying~each~time.\~\~Which~is~easier~to~write~down~using~an~exponent~of~\color{blue}r:\~~~~~~~~~~~~~~~~~~~~~~~ \bbox[10pt, black, border:2pt solid red]{\Large\color{white}{n × n × ... (r~times) = n^r}}}$