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Question Number 178054 by Acem last updated on 12/Oct/22

Mr. w wants to distribute n+1 different prizes to n friends so that each one gets at least one prize, how many results of this process. I hope to be that one gets 2

$${Mr}.\:{w}\:{wants}\:{to}\:{distribute}\:{n}+\mathrm{1}\:{different}\: \\ $$$${prizes}\:\:{to}\:{n}\:{friends}\:{so}\:{that}\:{each}\:{one}\:{gets} \\ $$$$\:{at}\:{least}\:{one}\:{prize},\:{how}\:{many}\:{results} \\ $$$$\:{of}\:{this}\:{process}.\:{I}\:{hope}\:{to}\:{be}\:{that}\:{one}\:{gets}\:\mathrm{2} \\ $$$$ \\ $$

Commented by mr W last updated on 12/Oct/22

wow! i′d like to really give a prize to each of you!

$${wow}!\:{i}'{d}\:{like}\:{to}\:{really}\:{give}\:{a}\:{prize}\:{to}\: \\ $$$${each}\:{of}\:{you}! \\ $$

Commented by mr W last updated on 12/Oct/22

my answer is ((n(n+1)!)/2).

$${my}\:{answer}\:{is}\:\frac{{n}\left({n}+\mathrm{1}\right)!}{\mathrm{2}}. \\ $$

Commented by Acem last updated on 12/Oct/22

In fact since the (n−1) are so much jealous i hardly got two.... Thx for the prizes The work was great! I wish you continued success

$${In}\:{fact}\:{since}\:{the}\:\left({n}−\mathrm{1}\right)\:{are}\:{so}\:{much}\:{jealous} \\ $$$$\:{i}\:{hardly}\:{got}\:{two}....\:{Thx}\:{for}\:{the}\:{prizes}\: \\ $$$$ \\ $$$${The}\:{work}\:{was}\:{great}!\:{I}\:{wish}\:{you} \\ $$$$\:{continued}\:{success} \\ $$

Answered by aurpeyz last updated on 12/Oct/22

first gift =n options second gift=n options third gift=(n−1) options fourth gift=(n−2) options fifth gift=(n−3) options sixth gift=(n−4) options ... (n)th gift=(n+1−4)options (n+1)th gift=(n+2−4)=(n−2) options ans=n×n×(n−1)×(n−2)×(n−3)× (n−4)×...×(n−2) I ended up confused. lol

$${first}\:{gift}\:={n}\:{options} \\ $$$${second}\:{gift}={n}\:{options} \\ $$$${third}\:{gift}=\left({n}−\mathrm{1}\right)\:{options} \\ $$$${fourth}\:{gift}=\left({n}−\mathrm{2}\right)\:{options} \\ $$$${fifth}\:{gift}=\left({n}−\mathrm{3}\right)\:{options} \\ $$$${sixth}\:{gift}=\left({n}−\mathrm{4}\right)\:{options} \\ $$$$... \\ $$$$\left({n}\right){th}\:{gift}=\left({n}+\mathrm{1}−\mathrm{4}\right){options} \\ $$$$\left({n}+\mathrm{1}\right){th}\:{gift}=\left({n}+\mathrm{2}−\mathrm{4}\right)=\left({n}−\mathrm{2}\right)\:{options} \\ $$$${ans}={n}×{n}×\left({n}−\mathrm{1}\right)×\left({n}−\mathrm{2}\right)×\left({n}−\mathrm{3}\right)× \\ $$$$\left({n}−\mathrm{4}\right)×...×\left({n}−\mathrm{2}\right) \\ $$$$ \\ $$$${I}\:{ended}\:{up}\:{confused}.\:{lol} \\ $$$$ \\ $$$$ \\ $$

Commented by Acem last updated on 12/Oct/22

The attempts are excellent even if it was not true cause it gaves a good figure for solving problems (:

$${The}\:{attempts}\:{are}\:{excellent}\:{even}\:{if}\:{it}\:{was}\:{not} \\ $$$$\:{true}\:{cause}\:{it}\:{gaves}\:{a}\:{good}\:{figure}\:{for}\:{solving} \\ $$$$\:{problems}\:\left(:\right. \\ $$

Commented by Acem last updated on 12/Oct/22

Let′s try now, There′s one friend_((?)) will get 2 differents_((?)) prizes that′s mean we′re in front of two operations of selection: (2 ways) Bring 2 diff.priz. (((And)),((( ×))) ) Select one fr._(to give) (((n+1)),(( 2)) ) × n = (((n+1)n^2 )/2) = NumWays_(2 pr. to 1fr.) Gifting n friends = gift_((n−1)pr. to(n−1)fr.) (((And)),(( (×))) ) gift_(2pr. to 1fr.) = P_(n−1) ^( n−1) × NumWays_(2 pr. to 1fr.) = (((n−1)!)/(0!)) × ((n.n.(n+1))/2)= ((n.n!.(n+1))/2) = ((n.(n+1)!)/2) ways

$${Let}'{s}\:{try}\:{now},\:{There}'{s}\:\boldsymbol{{one}}\:\boldsymbol{{friend}}_{\left(?\right)} \:{will} \\ $$$$\:{get}\:\mathrm{2}\:\boldsymbol{{differents}}_{\left(?\right)} \:\:{prizes} \\ $$$${that}'{s}\:{mean}\:{we}'{re}\:{in}\:{front}\:{of}\:{two}\:{operations} \\ $$$$\:{of}\:{selection}:\:\left(\mathrm{2}\:{ways}\right) \\ $$$$\:\underline{{Bring}}\:\mathrm{2}\:{diff}.{priz}.\:\begin{pmatrix}{\boldsymbol{{And}}}\\{\left(\:×\right)}\end{pmatrix}\:\underline{{Select}}\:{one}\:{fr}._{{to}\:{give}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\mathrm{2}}\end{pmatrix}\:×\:{n} \\ $$$$\:\:=\:\frac{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }{\mathrm{2}}\:=\:{NumWays}_{\mathrm{2}\:{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$ \\ $$$$\:{Gifting}\:{n}\:{friends}\:= \\ $$$$\:{gift}_{\left({n}−\mathrm{1}\right){pr}.\:{to}\left({n}−\mathrm{1}\right){fr}.} \:\begin{pmatrix}{\boldsymbol{{And}}}\\{\:\left(×\right)}\end{pmatrix}\:{gift}_{\mathrm{2}{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$ \\ $$$$\:=\:{P}_{{n}−\mathrm{1}} ^{\:{n}−\mathrm{1}} \:×\:{NumWays}_{\mathrm{2}\:{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$\:=\:\frac{\left({n}−\mathrm{1}\right)!}{\mathrm{0}!}\:×\:\frac{{n}.{n}.\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\:\frac{{n}.{n}!.\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:=\:\frac{{n}.\left({n}+\mathrm{1}\right)!}{\mathrm{2}}\:{ways} \\ $$$$\: \\ $$

Answered by mr W last updated on 12/Oct/22

method 1: i take a stranger from the street and bring him to my n friends. now i have n+1 “friends”. one of them is a fake friend. i give a prize to each of them, there are (n+1)! ways to do that. since the stranger is not a real friend, he should give his prize to one of my real friends. he has n ways to select a real friend. so i have n(n+1)! ways to distribute n+1 prizes to n friends. but since it′s the same whether some one gets at first the prize A and then from the stranger the prize B or whether he gets at first the prize B and then from the stranger the prize A. therefore in fact i have only ((n(n+1)!)/2) different ways.

$${method}\:\mathrm{1}: \\ $$$${i}\:{take}\:{a}\:{stranger}\:{from}\:{the}\:{street}\:{and} \\ $$$${bring}\:{him}\:{to}\:{my}\:{n}\:{friends}.\:{now}\:{i}\:{have} \\ $$$${n}+\mathrm{1}\:``{friends}''.\:{one}\:{of}\:{them}\:{is}\:{a}\:{fake} \\ $$$${friend}.\:{i}\:{give}\:{a}\:{prize}\:{to}\:{each}\:{of}\:{them}, \\ $$$${there}\:{are}\:\left({n}+\mathrm{1}\right)!\:{ways}\:{to}\:{do}\:{that}.\: \\ $$$${since}\:{the}\:{stranger}\:{is}\:{not}\:{a}\:{real}\:{friend}, \\ $$$${he}\:{should}\:{give}\:{his}\:{prize}\:{to}\:{one}\:{of}\:{my} \\ $$$${real}\:{friends}.\:{he}\:{has}\:{n}\:{ways}\:{to}\:{select} \\ $$$${a}\:{real}\:{friend}.\:{so}\:{i}\:{have}\:{n}\left({n}+\mathrm{1}\right)!\:{ways} \\ $$$${to}\:{distribute}\:{n}+\mathrm{1}\:{prizes}\:{to}\:{n}\:{friends}. \\ $$$${but}\:{since}\:{it}'{s}\:{the}\:{same}\:{whether}\:{some} \\ $$$${one}\:{gets}\:{at}\:{first}\:{the}\:{prize}\:{A}\:{and}\:{then}\: \\ $$$${from}\:{the}\:{stranger}\:{the}\:{prize}\:{B}\:{or}\: \\ $$$${whether}\:{he}\:{gets}\:{at}\:{first}\:{the}\:{prize}\:{B}\: \\ $$$${and}\:{then}\:{from}\:{the}\:{stranger}\:{the}\:{prize} \\ $$$${A}.\:{therefore}\:{in}\:{fact}\:{i}\:{have}\:{only}\: \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)!}{\mathrm{2}}\:{different}\:{ways}. \\ $$

Commented by mr W last updated on 12/Oct/22

method 2: one friend gets 2 prizes. to select this friend, there are n ways. he selects 2 from n+1 prizes, there are C_2 ^(n+1) =(((n+1)n)/2) ways. to distribute the remaining n−1 prizes among the n−1 friends, there are (n−1)! ways. ⇒n×(((n+1)n)/2)×(n−1)!=((n(n+1)!)/2)

$${method}\:\mathrm{2}: \\ $$$${one}\:{friend}\:{gets}\:\mathrm{2}\:{prizes}.\:{to}\:{select}\:{this} \\ $$$${friend},\:{there}\:{are}\:{n}\:{ways}. \\ $$$${he}\:{selects}\:\mathrm{2}\:{from}\:{n}+\mathrm{1}\:{prizes},\:{there} \\ $$$${are}\:{C}_{\mathrm{2}} ^{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right){n}}{\mathrm{2}}\:{ways}.\:{to}\:{distribute} \\ $$$${the}\:{remaining}\:{n}−\mathrm{1}\:{prizes}\:{among}\: \\ $$$${the}\:{n}−\mathrm{1}\:{friends},\:{there}\:{are}\:\left({n}−\mathrm{1}\right)!\: \\ $$$${ways}. \\ $$$$\Rightarrow{n}×\frac{\left({n}+\mathrm{1}\right){n}}{\mathrm{2}}×\left({n}−\mathrm{1}\right)!=\frac{{n}\left({n}+\mathrm{1}\right)!}{\mathrm{2}} \\ $$

Commented by Acem last updated on 12/Oct/22

The second method is the best! it′s like mine as the following: There′s one friend_((?)) will get 2 differents_((?)) prizes that′s mean we′re in front of two operations of selection: (2 ways) Bring 2 diff.priz. (((And)),((( ×))) ) Select one fr._(to give) (((n+1)),(( 2)) ) × n = (((n+1)n^2 )/2) = NumWays_(2 pr. to 1fr.) Gifting n friends = gift_((n−1)pr. to(n−1)fr.) (((And)),(( (×))) ) gift_(2pr. to 1fr.) = P_(n−1) ^( n−1) × NumWays_(2 pr. to 1fr.) = (((n−1)!)/(0!)) × ((n.n.(n+1))/2)= ((n.n!.(n+1))/2) = ((n.(n+1)!)/2) ways

$$\:{The}\:{second}\:{method}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{best}}!\:{it}'{s}\:{like}\:{mine} \\ $$$${as}\:{the}\:{following}: \\ $$$$ \\ $$$${There}'{s}\:\boldsymbol{{one}}\:\boldsymbol{{friend}}_{\left(?\right)} \:{will} \\ $$$$\:{get}\:\mathrm{2}\:\boldsymbol{{differents}}_{\left(?\right)} \:\:{prizes} \\ $$$${that}'{s}\:{mean}\:{we}'{re}\:{in}\:{front}\:{of}\:{two}\:{operations} \\ $$$$\:{of}\:{selection}:\:\left(\mathrm{2}\:{ways}\right) \\ $$$$\:\underline{{Bring}}\:\mathrm{2}\:{diff}.{priz}.\:\begin{pmatrix}{\boldsymbol{{And}}}\\{\left(\:×\right)}\end{pmatrix}\:\underline{{Select}}\:{one}\:{fr}._{{to}\:{give}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{pmatrix}{{n}+\mathrm{1}}\\{\:\:\:\mathrm{2}}\end{pmatrix}\:×\:{n} \\ $$$$\:\:=\:\frac{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }{\mathrm{2}}\:=\:{NumWays}_{\mathrm{2}\:{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$ \\ $$$$\:{Gifting}\:{n}\:{friends}\:= \\ $$$$\:{gift}_{\left({n}−\mathrm{1}\right){pr}.\:{to}\left({n}−\mathrm{1}\right){fr}.} \:\begin{pmatrix}{\boldsymbol{{And}}}\\{\:\left(×\right)}\end{pmatrix}\:{gift}_{\mathrm{2}{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$ \\ $$$$\:=\:{P}_{{n}−\mathrm{1}} ^{\:{n}−\mathrm{1}} \:×\:{NumWays}_{\mathrm{2}\:{pr}.\:{to}\:\mathrm{1}{fr}.} \\ $$$$\:=\:\frac{\left({n}−\mathrm{1}\right)!}{\mathrm{0}!}\:×\:\frac{{n}.{n}.\left({n}+\mathrm{1}\right)}{\mathrm{2}}=\:\frac{{n}.{n}!.\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:=\:\frac{{n}.\left({n}+\mathrm{1}\right)!}{\mathrm{2}}\:{ways} \\ $$$$\: \\ $$

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