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Question Number 35347 by chakraborty ankit last updated on 18/May/18

Answered by Rasheed.Sindhi last updated on 18/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{Let}{\mathrm{P}}_1,{\mathrm{P}}_2,{\mathrm{P}}_3,{\mathrm{P}}_4,{\mathrm{P}}_5\mathrm{\&}{\mathrm{P}}_6\mathrm{are}6\mathrm{vertices}.$$$${}^{\bullet }{\mathrm{P}}_1\mathrm{can}\mathrm{be}\mathrm{joined}\mathrm{to}5\mathrm{other}\mathrm{points}.\mathrm{Hence}$$$$\mathrm{producing}5\mathrm{straight}\mathrm{lines}.$$$${}^{\bullet }{\mathrm{P}}_2\mathrm{is}\mathrm{already}\mathrm{connected}\mathrm{to}{\mathrm{P}}_1.\mathrm{It}\mathrm{can}\mathrm{be}$$$$\mathrm{joined}4\mathrm{other}\mathrm{points}\mathrm{to}\mathrm{produce}4\mathrm{straight}$$$$\mathrm{lines}.$$$${}^{\bullet }{\mathrm{P}}_3\mathrm{is}\mathrm{to}\mathrm{join}3\mathrm{new}\mathrm{points}\mathrm{producing}$$$$3\mathrm{straight}\mathrm{lines}.$$$${}^{\bullet }{\mathrm{P}}_4\mathrm{produces}2\mathrm{new}\mathrm{straight}\mathrm{lines}$$$${}^{\bullet }{\mathrm{P}}_5:1\mathrm{straight}\mathrm{new}\mathrm{line}.$$$${}^{\bullet }{\mathrm{P}}_6:0\mathrm{new}\mathrm{straight}\mathrm{lines}.$$$$\mathcal{N}\mathrm{ote}:Thecentercannotproduceextra$$$$lineas{it}^{\prime }spartofsomeexistinglines$$$$\mathrm{Total}\mathrm{straight}\mathrm{lines}:5+4+3+2+1+0$$$$=15$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{There}\mathrm{are}\mathrm{total}7\mathrm{points}.$$$$\mathrm{Number}\mathrm{of}\mathrm{triplets}=\left(\begin{array}{c}7\\ 3\end{array}\right)=\frac{7.6.5}{1.2.3}=35$$$$3\mathrm{triplets}\mathrm{are}\mathrm{not}\mathrm{traingles}.\mathrm{They}\mathrm{are}$$$$\mathrm{straight}\mathrm{lines}.$$$$\mathrm{Hence}\mathrm{number}\mathrm{of}\mathrm{triangles}=35-3=32$$

Commented by Rasheed.Sindhi last updated on 18/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{way}.$$$${}^{\bullet }\mathrm{There}\mathrm{are}7\mathrm{points},\mathrm{out}\mathrm{of}\mathrm{them}\mathrm{only}6\mathrm{vertices}$$$$\mathrm{are}\mathrm{capable}\mathrm{to}\mathrm{produce}\mathrm{all}\mathrm{possible}\mathrm{lines}.\mathrm{Any}\mathrm{pair}$$$$\mathrm{of}\mathrm{vertices}\mathrm{can}\mathrm{produce}\mathrm{a}\mathrm{straight}\mathrm{line}$$$$\mathrm{So}\mathrm{number}\mathrm{of}\mathrm{straight}\mathrm{lines}$$$$=\mathrm{number}\mathrm{of}\mathrm{pairs}\mathrm{of}\mathrm{vertices}$$$$\mathrm{Number}\mathrm{of}\mathrm{pairs}=\left(\begin{array}{c}6\\ 2\end{array}\right)=\frac{6.5}{1.2}=15$$$$\mathrm{Or}\mathrm{number}\mathrm{of}\mathrm{straight}\mathrm{lines}=15$$

Answered by Rasheed.Sindhi last updated on 18/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$★\mathrm{Join}\mathrm{of}\mathrm{two}\mathrm{non}-\mathrm{consecutive}\mathrm{vertices}$$$$\mathrm{is}\mathrm{defined}\mathrm{as}\mathbf{diagonal}\mathrm{of}\mathrm{polygon}.$$$$★\mathrm{Number}\mathrm{of}\mathrm{non}-\mathrm{consecutive}\mathrm{vertices}$$$$\mathrm{to}\mathrm{a}\mathrm{particular}\mathrm{vertex}\mathrm{in}\mathrm{an}\mathrm{n}-\mathrm{gon}\mathrm{is}$$$$\mathrm{n}-3\left(3\mathrm{points}\mathrm{are}\mathrm{consecutive}\right)$$$$\mathrm{So}\mathrm{in}\mathrm{octagon}\mathrm{to}\mathrm{a}\mathrm{particular}\mathrm{vertex}$$$$8-2=6\mathrm{non}-\mathrm{consecutive}\mathrm{vertices}.$$$$\mathrm{Let}\mathrm{the}\mathrm{vertices}\mathrm{are}{\mathrm{P}}_1,{\mathrm{P}}_2,...,{\mathrm{P}}_8$$$$\divideontimes \mathrm{Begin}\mathrm{with}{\mathrm{P}}_1:{\mathrm{P}}_1\mathrm{can}\mathrm{be}\mathrm{joined}{\mathrm{P}}_3,...,{\mathrm{P}}_7$$$$\mathrm{in}\mathrm{order}\mathrm{to}\mathrm{produce}5\mathrm{diagonals}.$$$$(\mathrm{Of}\mathrm{course}{\mathrm{P}}_1{\mathrm{P}}_2\mathrm{\&}{\mathrm{P}}_1{\mathrm{P}}_8\mathrm{are}\mathrm{not}\mathrm{diagonals}$$$$\mathrm{because}{\mathrm{P}}_2\mathrm{\&}{\mathrm{P}}_8\mathrm{are}\mathrm{consecutive}\mathrm{to}{\mathrm{P}}_1)$$$$\divideontimes \mathrm{Similarly}{\mathrm{P}}_2\mathrm{can}\mathrm{be}\mathrm{joined}\mathrm{all}\mathrm{other}$$$$\mathrm{points}\mathrm{except}{\mathrm{P}}_1\mathrm{\&}{\mathrm{P}}_3\mathrm{to}\mathrm{produce}5\mathrm{new}$$$$\mathrm{diagonals}.$$$$\divideontimes {\mathrm{P}}_3,\mathrm{however}\mathrm{already}\mathrm{conected}\mathrm{to}{\mathrm{P}}_1$$$$\mathrm{so}\mathrm{the}\mathrm{diagonal}{\mathrm{P}}_3{\mathrm{P}}_1\mathrm{is}\mathrm{already}\mathrm{counted}.$$$$\mathrm{Hence}\mathrm{it}\mathrm{produces}4\mathrm{new}\mathrm{diagonals}.$$$$\divideontimes \mathrm{In}\mathrm{this}\mathrm{way}$$$${\mathrm{P}}_4\mathrm{produces}3\mathrm{new}\mathrm{diagonals}.$$$${\mathrm{P}}_5\mathrm{produces}2\mathrm{new}\mathrm{diagonals}.$$$${\mathrm{P}}_6\mathrm{produces}1\mathrm{new}\mathrm{diagonals}.$$$${\mathrm{P}}_7\mathrm{\&}{\mathrm{P}}_8\mathrm{produce}\mathrm{no}\mathrm{new}\mathrm{diagonal}.$$$$\mathrm{Total}\mathrm{of}\mathrm{diagonals}=5+5+4+3+2+1$$$$=20$$

Answered by Rasheed.Sindhi last updated on 19/May/18

$$\text{You can't use 'macro parameter character \#' in math mode}$$$${}^{\bullet }3\mathrm{collinear}\mathrm{points}\mathrm{produce}1\mathrm{line}.$$$$\mathrm{Each}\mathrm{of}\mathrm{three}\mathrm{points}\mathrm{can}\mathrm{be}\mathrm{joined}\mathrm{to}$$$$4+5=9\mathrm{points}$$$$\mathrm{Each}\mathrm{point}\mathrm{produce}9\mathrm{lines}$$$$\mathrm{All}3\mathrm{points}\mathrm{produce}9\times 3=27\mathrm{lines}.$$$${}^{\bullet }4\mathrm{collinear}\mathrm{points}\mathrm{produce}1\mathrm{line}.$$$$\mathrm{Each}\mathrm{of}4\mathrm{points}\mathrm{can}\mathrm{be}\mathrm{joined}\mathrm{to}$$$$5\mathrm{points}\mathrm{to}\mathrm{produce}\mathbf{new}\mathrm{lines}$$$$\mathrm{Each}\mathrm{point}\mathrm{produce}5\mathrm{lines}.$$$$\mathrm{All}4\mathrm{points}\mathrm{produce}5\times 4=20\mathbf{new}\mathrm{lines}.$$$${}^{\bullet }5\mathrm{collinear}\mathrm{points}\mathrm{produce}1\mathrm{line}.$$$$\mathrm{All}\mathrm{the}\mathrm{lines}\mathrm{produced}\mathrm{by}\mathrm{these}$$$$\mathrm{points}\mathrm{have}\mathrm{already}\mathrm{been}\mathrm{counted}.$$$$\mathrm{So}\mathrm{produce}0\mathbf{new}\mathrm{line}.$$$$\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{lines}=$$$$1+27+1+20+1=50$$$$$$

Commented by chakraborty ankit last updated on 21/May/18

$$thankalotsir$$

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