tag:blogger.com,1999:blog-2514352312859447561.post1495397483386085905..comments2024-01-30T03:30:45.740-05:00Comments on Bad Data, Bad!: Wednesday Brain Teaser 2-6-12bs kinghttp://www.blogger.com/profile/02871717971078952304noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-2514352312859447561.post-64984329297081217492013-02-10T00:13:51.062-05:002013-02-10T00:13:51.062-05:00Well, sometime after I admitted inability to think...Well, sometime after I admitted inability to think my way through it (at that moment), I hit a search engine. <br /><br />And found <a href="https://en.wikipedia.org/wiki/Derangement" rel="nofollow">that formula</a>. Plus a recursive one that I like better.<br /><br />But I figured I would wait, and see if anyone else got close.<br /><br />This was definitely a challenging puzzle, but now I think that I should have been able to solve it without a search engine...SJhttps://www.blogger.com/profile/12043843405366080460noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-73324565094814006822013-02-07T21:08:40.603-05:002013-02-07T21:08:40.603-05:00I would never have gotten there.I would never have gotten there.Assistant Village Idiothttps://www.blogger.com/profile/01978011985085795099noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-47841157103862898762013-02-07T20:13:17.077-05:002013-02-07T20:13:17.077-05:00And the prize goes to James!
In case you're c...And the prize goes to James!<br /><br />In case you're curious, the equation for this is <br />n!((1/(2!)-1/(3!)+1/(4!)-1/(5!)+1/(6!)-1/(7!))bs kinghttps://www.blogger.com/profile/02871717971078952304noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-8336024056749819962013-02-07T19:12:47.132-05:002013-02-07T19:12:47.132-05:00A program to count the permutations also gives 185...A program to count the permutations also gives 1854.jameshttps://www.blogger.com/profile/01792036361407527304noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-51232085127737512292013-02-07T18:47:20.423-05:002013-02-07T18:47:20.423-05:00I tried this, and maybe the sequence is instructiv...I tried this, and maybe the sequence is instructive:<br />6 cases for the first, 5 for the second, 4 .. Wait a bit--suppose the first and second just swapped? Start from the beginning.<br /><br />1 letter, always right, =0<br />2 letters, one way to goober it up<br />3 letters, 2 ways<br />4 letters: ah. 3*2=6 ways of cycling the 4 around, but then 3 ways of doing 2+2. (=9)<br /><br />So partitioning is the way to go. A partition of 1 always maps the letter into the right envelope, so there's no answers with partition 1.<br /><br />7 can be<br /> a "cycle" of 7: with 1 case, which has 6*5*4*3*2 permutations<br /> a "cycle" of 5 plus a "cycle" of 2: with 7*6/2 cases and 4*3*2 permutations<br /> a "cycle" of 4 plus a "cycle" of 3: where I'm going to cheat and include the 2+2 in the 4, with 7*6*5/6 cases and (2 {for the 3} * 9 {for the 4}) permutations<br /><br />That gives me 720+504+630=1854jameshttps://www.blogger.com/profile/01792036361407527304noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-43076278838997431332013-02-07T13:51:18.715-05:002013-02-07T13:51:18.715-05:00I don't think it's that clean. I'm fig...I don't think it's that clean. I'm figuring 7! minus a whole bunch of werd stuff is going to show up somewhere, and that will be much larger.<br /><br />For one letter there are 0 wrong arrangements. For 2 letters, one wrong arrangement. For 3 letters, 2 wrongs. For 4 letters, 9 possible wrong arrangements. I started working 5 by hand, and the number is over 40, but my head hurts and I may have missed something.<br /><br />So what's got 5!(120) minus stuff, or 4! (24) plus stuff, and comes out to something around 45? Dunno. It must be a series, so I can't intuit it in from there. I'm letting this spin for qwhile.Assistant Village Idiothttps://www.blogger.com/profile/01978011985085795099noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-10308018777452519522013-02-07T11:33:41.241-05:002013-02-07T11:33:41.241-05:00Letters: A B C D E F G
Envelopes 1 2 3 4 5 6 7
...Letters: A B C D E F G<br />Envelopes 1 2 3 4 5 6 7 <br /><br />A can be put incorrectly in 2-7, so there's 6 options there.<br /><br />B can be put in 1, 3-7 (i.e. 6 options), if A was put in 2, Otherwise, there are 5 options<br /><br />If A or B was put in 3, C can go in 5 places. Otherwise, 4 places<br /><br />If A, B or C was put in 4, D can go in 4 places, otherwise 3<br /><br />If A,B,C or D were put in 5, E can go in 3 places, otherwise 2<br /><br />If any of A-E were put in 6, F can go in 2 places, otherwise 1<br /><br />G can go in 1 place<br /><br /><br />I am running short on time, but I would guess 6x5x4x3x2 + 6+5+4+3+2 -> 740 different combinations<br /><br /><br />Geek Vaderhttps://www.blogger.com/profile/05729378883810714043noreply@blogger.comtag:blogger.com,1999:blog-2514352312859447561.post-51849428506217401722013-02-07T08:24:53.506-05:002013-02-07T08:24:53.506-05:00I should know the answer to this one...or to be ab...I should know the answer to this one...or to be able to figure it out.<br /><br />But I can't think of it right now.SJhttps://www.blogger.com/profile/12043843405366080460noreply@blogger.com