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Question Number 117286 by Dwaipayan Shikari last updated on 10/Oct/20

$$Apersonwantstoinvitehis6friendsinaDinnerparty.$$$$Hehas3persontosendlettertothem.Inhowmanyways$$$$hecaninvitehis6friends?$$

Answered by mr W last updated on 10/Oct/20

$$pleaseexplainwhatyoumeanwith$$$$‘‘Hehas3persontosendletterto{them}^{″}.$$

Commented by Dwaipayan Shikari last updated on 10/Oct/20

$$3postman,And6freinds$$

Commented by TANMAY PANACEA last updated on 10/Oct/20

$$plsconveytoadminthatican[notanswer...itcollapsepls$$

Commented by mr W last updated on 10/Oct/20

$$toplacendistinctobjectsintok$$$$distinctboxesthereare{k}^{n}ways.$$$$\Rightarrow answeris{3}^6=729$$

Commented by Tinku Tara last updated on 10/Oct/20

$$\mathrm{Dear}\mathrm{Tanmay},\mathrm{a}\mathrm{fix}\mathrm{for}\mathrm{the}\mathrm{issue}$$$$\mathrm{while}\mathrm{answering}\mathrm{post}\mathrm{has}\mathrm{been}\mathrm{released}.$$$$\mathrm{Are}\mathrm{u}\mathrm{still}\mathrm{facing}\mathrm{problems}?$$

Commented by TANMAY PANACEA last updated on 10/Oct/20

$$problemsolved..thankyou$$

Answered by PRITHWISH SEN 2 last updated on 10/Oct/20

$$3^6=729\mathrm{ways}$$$$\mathrm{let}\mathrm{if}\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{pos}\stackrel{}{\mathrm{t}man}\mathbf{a}\mathrm{and}\mathbf{b}$$$$\mathrm{and}\mathrm{they}\mathrm{have}\mathrm{to}\mathrm{go}\mathrm{to}4\mathrm{places}$$$$\mathrm{then}$$$${(\mathrm{a}+\mathrm{b})}^4={\mathrm{a}}^4+{4\mathrm{a}}^3\mathrm{b}+{6\mathrm{a}}^2{\mathrm{b}}^2+{4\mathrm{ab}}^3+{\mathrm{b}}^4$$$$\mathrm{i}.\mathrm{e}$$$$\mathbf{a}\mathrm{has}\mathrm{gone}\mathrm{in}4\mathrm{places}\mathrm{in}1\mathrm{way}$$$$\mathbf{a}\mathrm{has}\mathrm{gone}\mathrm{in}3\mathrm{places}\mathrm{and}\mathbf{b}\mathrm{in}1\mathrm{in}4\mathrm{ways}$$$$\mathbf{a}\mathrm{has}\mathrm{gone}\mathrm{in}2\mathrm{places}\mathrm{and}\mathbf{b}\mathrm{in}2\mathrm{in}6\mathrm{ways}$$$$\mathbf{a}\mathrm{has}\mathrm{gone}\mathrm{in}1\mathrm{places}\mathrm{and}\mathbf{b}\mathrm{in}3\mathrm{in}4\mathrm{ways}$$$$\mathbf{b}\mathrm{has}\mathrm{gone}\mathrm{in}4\mathrm{places}\mathrm{in}1\mathrm{way}$$$$\mathrm{i}.\mathrm{e}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{co}\mathrm{efficient}=1+4+6+4+1=16$$$$\mathrm{or}\mathrm{by}\mathrm{putting}\mathrm{a}=\mathrm{b}=\mathrm{l}{(1+1)}^4=2^4=16\mathrm{ways}$$$$\mathrm{now}\mathrm{for}\mathrm{this}\mathrm{problem}$$$$\mathrm{on}{(\mathrm{a}+\mathrm{b}+\mathrm{c})}^6\mathrm{putting}\mathrm{a}=\mathrm{b}=\mathrm{c}=1\mathrm{we}\mathrm{get}{3}^6\mathrm{ways}$$

Commented by Dwaipayan Shikari last updated on 10/Oct/20

$$Thankingallofyou$$

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