tag:blogger.com,1999:blog-38193052.post8624882538959180703..comments2024-03-20T08:57:17.447-03:00Comments on Jornalheiros: Amigo ocultoPC Filhohttp://www.blogger.com/profile/16547063456626761789noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-38193052.post-68391487568506929562015-11-01T11:11:26.931-02:002015-11-01T11:11:26.931-02:00And yes, this problem is very tricky. Most people ...And yes, this problem is very tricky. Most people answer 120...PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-44429163510123569172015-11-01T11:09:48.998-02:002015-11-01T11:09:48.998-02:00Perfeito, Marcelo! Há 265 maneiras possíveis de tr...Perfeito, Marcelo! Há 265 maneiras possíveis de trocar os presentes. :-)PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-73469734793618210592015-11-01T09:25:59.460-02:002015-11-01T09:25:59.460-02:00 PS-se não fosse a sua observação, eu acharia que ... PS-se não fosse a sua observação, eu acharia que o Jake estaria corretoAnonymoushttps://www.blogger.com/profile/14266294534241424190noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-10672332812982110482015-11-01T09:20:23.795-02:002015-11-01T09:20:23.795-02:00Caro PC,
Seja A[N] as combinações de distribuição...Caro PC, <br />Seja A[N] as combinações de distribuição de presentes para N amigos ocultos.<br />Se o primeiro sujeito A selecionar B e depois B devolver para A, temos o problema resumido a A[N-2] ( o mesmo problema sem A nem B).<br />Se B entregar para outro amigo, o problema se resume a A[N-1].<br />Como A pode selecionar N-1 amigos ocultos, temos A[N]=(N-1)*(A[N-1]+A[N-2])<br />Então A[6]= 5*(A[5]+A[4])<br />A[4]=9 (você mostrou)<br />A[5] =4*(A[4]+A[3])=4*(9+2)=44<br />Voltando para A[6], Temos:<br />A[6]=5*(44+9)=265<br />ST<br />Ps- se mão fosse a sua observação, eu acharia que o Jake está correto<br />Anonymoushttps://www.blogger.com/profile/14266294534241424190noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-32193097524129511432015-10-31T17:59:51.357-02:002015-10-31T17:59:51.357-02:00Jake, 120 is not the correct answer.
To explain w...Jake, 120 is not the correct answer.<br /><br />To explain why, let's suppose only 4 people participate in the play. According to your reasoning, the number of possible scenarios would be 3! = 6.<br /><br />However, there are 9 possible scenarios in fact:<br />1) A gives to B, B gives to C, C gives to D, D gives to A.<br />2) A gives to B, B gives to D, D gives to C, C gives to A.<br />3) A gives to B, B gives to A, C gives to D, D gives to C.<br />4) A gives to C, C gives to B, B gives to D, D gives to A.<br />5) A gives to C, C gives to D, D gives to B, B gives to A.<br />6) A gives to C, C gives to A, B gives to D, D gives to B.<br />7) A gives to D, D gives to B, B gives to C, C gives to A.<br />8) A gives to D, D gives to C, C gives to B, B gives to A.<br />9) A gives to D, D gives to A, B gives to C, C gives to B.PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-51167255705425495652015-10-31T17:37:40.038-02:002015-10-31T17:37:40.038-02:00For simplicity's sake, let's assume that t...For simplicity's sake, let's assume that the people pick the gifts sequentially. The first person to pick would have five possibilities to choose from, then the second person would have four, etc. Therefore, the total number of possible gifting scenarios is 5! = 120.<br /><br />My answer: 120jrh150482https://www.blogger.com/profile/10502831081969372299noreply@blogger.com