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Question Number 117286 by Dwaipayan Shikari last updated on 10/Oct/20
A person wants to invite his 6 friends in a Dinner party.  He has 3 person to send letter to them.In how many ways  he can invite his 6 friends?
$${A}\:{person}\:{wants}\:{to}\:{invite}\:{his}\:\mathrm{6}\:{friends}\:{in}\:{a}\:{Dinner}\:{party}. \\ $$$${He}\:{has}\:\mathrm{3}\:{person}\:{to}\:{send}\:{letter}\:{to}\:{them}.{In}\:{how}\:{many}\:{ways} \\ $$$${he}\:{can}\:{invite}\:{his}\:\mathrm{6}\:{friends}? \\ $$
Answered by mr W last updated on 10/Oct/20
please explain what you mean with  “He has 3 person to send letter to them”.
$${please}\:{explain}\:{what}\:{you}\:{mean}\:{with} \\ $$$$“{He}\:{has}\:\mathrm{3}\:{person}\:{to}\:{send}\:{letter}\:{to}\:{them}''. \\ $$
Commented by Dwaipayan Shikari last updated on 10/Oct/20
3 postman , And  6 freinds
$$\mathrm{3}\:{postman}\:,\:{And}\:\:\mathrm{6}\:{freinds} \\ $$
Commented by TANMAY PANACEA last updated on 10/Oct/20
pls convey to admin that i can[not answer...it collapse pls
$${pls}\:{convey}\:{to}\:{admin}\:{that}\:{i}\:{can}\left[{not}\:{answer}…{it}\:{collapse}\:{pls}\right. \\ $$
Commented by mr W last updated on 10/Oct/20
to place n distinct objects into k  distinct boxes there are k^n  ways.  ⇒answer is 3^6 =729
$${to}\:{place}\:{n}\:{distinct}\:{objects}\:{into}\:{k} \\ $$$${distinct}\:{boxes}\:{there}\:{are}\:{k}^{{n}} \:{ways}. \\ $$$$\Rightarrow{answer}\:{is}\:\mathrm{3}^{\mathrm{6}} =\mathrm{729} \\ $$
Commented by Tinku Tara last updated on 10/Oct/20
Dear Tanmay, a fix for the issue  while answering post has been released.  Are u still facing problems?
$$\mathrm{Dear}\:\mathrm{Tanmay},\:\mathrm{a}\:\mathrm{fix}\:\mathrm{for}\:\mathrm{the}\:\mathrm{issue} \\ $$$$\mathrm{while}\:\mathrm{answering}\:\mathrm{post}\:\mathrm{has}\:\mathrm{been}\:\mathrm{released}. \\ $$$$\mathrm{Are}\:\mathrm{u}\:\mathrm{still}\:\mathrm{facing}\:\mathrm{problems}? \\ $$
Commented by TANMAY PANACEA last updated on 10/Oct/20
problem solved ..thank you
$${problem}\:{solved}\:..{thank}\:{you} \\ $$
Answered by PRITHWISH SEN 2 last updated on 10/Oct/20
3^6  = 729 ways  let if there are two post^ man a and b  and they have to go to 4 places  then   (a+b)^4  = a^4 +4a^3 b+6a^2 b^2 +4ab^3 +b^4   i.e  a has gone in 4 places in 1 way  a has gone in 3 places andb in 1 in 4 ways  a has gone in 2 places and b in 2 in 6 ways  a has gone in 1 places and b in 3 in 4 ways  b has gone in 4 places in 1 way  i.e sum of the co efficient = 1+4+6+4+1=16  or by putting a=b=l    (1+1)^4 =2^4 =16 ways  now for this problem  on (a+b+c)^6   putting a=b=c=1 we get 3^6 ways
$$\mathrm{3}^{\mathrm{6}} \:=\:\mathrm{729}\:\mathrm{ways} \\ $$$$\mathrm{let}\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{two}\:\mathrm{pos}\overset{} {\mathrm{t}man}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{and}\:\mathrm{they}\:\mathrm{have}\:\mathrm{to}\:\mathrm{go}\:\mathrm{to}\:\mathrm{4}\:\mathrm{places} \\ $$$$\mathrm{then}\: \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{4}} \:=\:\mathrm{a}^{\mathrm{4}} +\mathrm{4a}^{\mathrm{3}} \mathrm{b}+\mathrm{6a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} +\mathrm{4ab}^{\mathrm{3}} +\mathrm{b}^{\mathrm{4}} \\ $$$$\mathrm{i}.\mathrm{e} \\ $$$$\boldsymbol{\mathrm{a}}\:\mathrm{has}\:\mathrm{gone}\:\mathrm{in}\:\mathrm{4}\:\mathrm{places}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way} \\ $$$$\boldsymbol{\mathrm{a}}\:\mathrm{has}\:\mathrm{gone}\:\mathrm{in}\:\mathrm{3}\:\mathrm{places}\:\mathrm{and}\boldsymbol{\mathrm{b}}\:\mathrm{in}\:\mathrm{1}\:\mathrm{in}\:\mathrm{4}\:\mathrm{ways} \\ $$$$\boldsymbol{\mathrm{a}}\:\mathrm{has}\:\mathrm{gone}\:\mathrm{in}\:\mathrm{2}\:\mathrm{places}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{in}\:\mathrm{2}\:\mathrm{in}\:\mathrm{6}\:\mathrm{ways} \\ $$$$\boldsymbol{\mathrm{a}}\:\mathrm{has}\:\mathrm{gone}\:\mathrm{in}\:\mathrm{1}\:\mathrm{places}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{in}\:\mathrm{3}\:\mathrm{in}\:\mathrm{4}\:\mathrm{ways} \\ $$$$\boldsymbol{\mathrm{b}}\:\mathrm{has}\:\mathrm{gone}\:\mathrm{in}\:\mathrm{4}\:\mathrm{places}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way} \\ $$$$\mathrm{i}.\mathrm{e}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{co}\:\mathrm{efficient}\:=\:\mathrm{1}+\mathrm{4}+\mathrm{6}+\mathrm{4}+\mathrm{1}=\mathrm{16} \\ $$$$\mathrm{or}\:\mathrm{by}\:\mathrm{putting}\:\mathrm{a}=\mathrm{b}=\mathrm{l}\:\:\:\:\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{4}} =\mathrm{2}^{\mathrm{4}} =\mathrm{16}\:\mathrm{ways} \\ $$$$\mathrm{now}\:\mathrm{for}\:\mathrm{this}\:\mathrm{problem} \\ $$$$\mathrm{on}\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{6}} \:\:\mathrm{putting}\:\mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{3}^{\mathrm{6}} \mathrm{ways} \\ $$
Commented by Dwaipayan Shikari last updated on 10/Oct/20
Thanking all of you
$${Thanking}\:{all}\:{of}\:{you} \\ $$

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