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Question-189716




Question Number 189716 by Mastermind last updated on 20/Mar/23
Answered by mr W last updated on 21/Mar/23
5!×2!=240
$$\mathrm{5}!×\mathrm{2}!=\mathrm{240} \\ $$
Commented by Mastermind last updated on 21/Mar/23
Someone said its 48
$$\mathrm{Someone}\:\mathrm{said}\:\mathrm{its}\:\mathrm{48} \\ $$
Commented by mr W last updated on 21/Mar/23
what do you say?
$${what}\:{do}\:{you}\:{say}? \\ $$
Commented by mr W last updated on 21/Mar/23
method 1:  at first only the 5 people of the friend  family will be seated. there are 4!  ways to do this. betweem there are  5 places where peter and his wife  can be seated. peter and his wife can  change their seats, there are 2 ways.  so totally there are  4!×5×2=240 ways.    method 2:  at first only peter and the 5 people  from the friend family will be seated.   to arrange the 6 people at the table  there are 5! ways. now peter′s wife  takes a seat next to peter, right to  him or left to him, so 2 ways.   totally 5!×2=240 ways.
$${method}\:\mathrm{1}: \\ $$$${at}\:{first}\:{only}\:{the}\:\mathrm{5}\:{people}\:{of}\:{the}\:{friend} \\ $$$${family}\:{will}\:{be}\:{seated}.\:{there}\:{are}\:\mathrm{4}! \\ $$$${ways}\:{to}\:{do}\:{this}.\:{betweem}\:{there}\:{are} \\ $$$$\mathrm{5}\:{places}\:{where}\:{peter}\:{and}\:{his}\:{wife} \\ $$$${can}\:{be}\:{seated}.\:{peter}\:{and}\:{his}\:{wife}\:{can} \\ $$$${change}\:{their}\:{seats},\:{there}\:{are}\:\mathrm{2}\:{ways}. \\ $$$${so}\:{totally}\:{there}\:{are} \\ $$$$\mathrm{4}!×\mathrm{5}×\mathrm{2}=\mathrm{240}\:{ways}. \\ $$$$ \\ $$$${method}\:\mathrm{2}: \\ $$$${at}\:{first}\:{only}\:{peter}\:{and}\:{the}\:\mathrm{5}\:{people} \\ $$$${from}\:{the}\:{friend}\:{family}\:{will}\:{be}\:{seated}.\: \\ $$$${to}\:{arrange}\:{the}\:\mathrm{6}\:{people}\:{at}\:{the}\:{table} \\ $$$${there}\:{are}\:\mathrm{5}!\:{ways}.\:{now}\:{peter}'{s}\:{wife} \\ $$$${takes}\:{a}\:{seat}\:{next}\:{to}\:{peter},\:{right}\:{to} \\ $$$${him}\:{or}\:{left}\:{to}\:{him},\:{so}\:\mathrm{2}\:{ways}.\: \\ $$$${totally}\:\mathrm{5}!×\mathrm{2}=\mathrm{240}\:{ways}. \\ $$
Answered by manolex last updated on 21/Mar/23
P_c =(n−1)!  Peter  and  wife             1     2!  anfitrion and wife      1      established sites  1!  3 family                             3  n=5  P_c =4!2!  P_c =48
$${P}_{{c}} =\left({n}−\mathrm{1}\right)! \\ $$$${Peter}\:\:{and}\:\:{wife}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}! \\ $$$${anfitrion}\:{and}\:{wife}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:{established}\:{sites}\:\:\mathrm{1}! \\ $$$$\mathrm{3}\:{family}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3} \\ $$$${n}=\mathrm{5} \\ $$$${P}_{{c}} =\mathrm{4}!\mathrm{2}! \\ $$$${P}_{{c}} =\mathrm{48} \\ $$$$ \\ $$
Commented by mr W last updated on 21/Mar/23
do peter and his wife belong to the  family which invited them? when   it′s a family of 5 members, how   many people totally are taking the   dinner?  A) 5      B) 7      C) none of both
$${do}\:{peter}\:{and}\:{his}\:{wife}\:{belong}\:{to}\:{the} \\ $$$${family}\:{which}\:{invited}\:{them}?\:{when}\: \\ $$$${it}'{s}\:{a}\:{family}\:{of}\:\mathrm{5}\:{members},\:{how}\: \\ $$$${many}\:{people}\:{totally}\:{are}\:{taking}\:{the}\: \\ $$$${dinner}? \\ $$$$\left.{A}\left.\right)\left.\:\mathrm{5}\:\:\:\:\:\:{B}\right)\:\mathrm{7}\:\:\:\:\:\:{C}\right)\:{none}\:{of}\:{both} \\ $$
Commented by manolex last updated on 21/Mar/23
$$ \\ $$
Commented by manolex last updated on 21/Mar/23
Answered by mehdee42 last updated on 21/Mar/23
case1)if thay are in a row ⇒ 6!×2!  case2) if they  are around a tsble ⇒ 5!×2!
$$\left.{case}\mathrm{1}\right){if}\:{thay}\:{are}\:{in}\:{a}\:{row}\:\Rightarrow\:\mathrm{6}!×\mathrm{2}! \\ $$$$\left.{case}\mathrm{2}\right)\:{if}\:{they}\:\:{are}\:{around}\:{a}\:{tsble}\:\Rightarrow\:\mathrm{5}!×\mathrm{2}! \\ $$

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