Can you solve: In how many different ways can two 8-tentacled octopi kiss? How did you get your answer.?
356432
hmmm... how would octupi kiss?\nIf they did it by touching tentacles, then there'd be 64 different ways (8^2), or if they touched all 8 tentacles, then there'd be 40320 different ways (8!).....(assuming they could tie themselves up a little bit!)...\nOr, maybe like the person above suggested, there's just one way...beak to beak!
156461
>>>>how many legs of a centipede. \n\nI think you meant to ask "How many legs HAS a centipede?". Actually given your grasp of grammar you probably meant to ask "How many legs does a centipede have?".\n\n\nThe answer is that it varies, anywhere from eight to over a hundred depending on the species.\n\n>>>>>>>Are the number of legs for a crab is always constant?\n\nI think you meant to ask "Are the number of legs for a crab always constant" or "Is the number of legs for a crab always constant".\n\nThe answer is that, barring accidental amputation, all ADULT crabs have 10 legs. Many people believe they have 8 legs, but crabs also have a pair of swimming legs that they keep tucked against the body when they are on land.\n\nJuvenile crabs can have just 6 legs.
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
94206
11.\nthe =3\nalphabet =8
925922
put two 2s facing each other and you will get a tiny fish ... yes you are correct :)
233027
There are 8! different arrangements (permutations) of the 8 friends.\n\nThere are 4! different arrangements of the 4 boys.\n\nThere are 5 different locations where the leftmost boy can sit when the 4 boys sit together.\n\nTherefore, there are 5*4! (or 5!) arrangements of the boys where all the boys sit together.  But there are also 4! arrangements of the 4 girls in the remaining slots.  So there are 5!*4! arrangements where the 4 boys sit together.\n\nSince there are 8! arrangements overall, the probability of all boys sitting together is (5!*4!)/8!.\n\nSimilarly, if the boys sit to the left and the girls sit to the right, then the probability is (4!*4!)/8!.\n\nInterestingly, this last answer is also the same probability as boy-girl-boy-girl-boy-girl-boy-girl and also the probability of girl-boy-girl-boy-girl-boy-girl-boy.  Therefore, the probability of alternating seats is (2*4!*4!)/8!.\n\nRecall that n! (read n-factorial) is n*(n-1)*(n-2)*...*3*2*1.
113373
haha u stoel this idea from that other lady nick...its probably  2 or 3 or 4 or 5 or 6 or 7 or 8 but im not gonna go with 9 on this one...
342259
6! or 720 arrangements.\n\n2 * 4! or 48 with vowels at the ends, so it's a 1/15 chance.\n\nEdit: Yeah, I messed up.  Forgot the *2.
1077026
Hmm.  Let me seeee.  <unwraps tootsie pop> \n\nA one........\n\nA two.......\n\nA three (CHOMP!!) <Bites tootsie pop>\n\nThree.
30747
2 cubes have 16 corners (since each cube has 8 corners)\n\na cylinder doesn't have any corners
934198
UH-UHH!   It's  28.  \n\nThe first one  we count has 7 distinct ribbons. One each to every other diver. The second one has 6 since the one to the first has already been counted. The next one has 5 since the ribbons to the previous 2 have already been counted and so on. \n  So the answer is 7+6+5+4+3+2+1= 28.  Notice that we only\naccount for 7 sky divers. That's because the 8th one's ribbons have all been accounted for by the others.
797253
12,000,198 microscopical creatures\n\nit's 7 x 7 x 7 x 7 x 17 x 7 x 6 x 7 = 12,000,198\n\nBy the way, why is this in Words and Wordplay not Math?\nAm I missing something?
1173841
The correct answer is two\n\nYoda and Luminara Undulli
1006024
According to MY dictionary (the Oxford pocket Dictionary and Thesaurus,) it's Octopuses.\n\nBut according to\nhttp://dictionary.reference.com/browse/octopi\noctopi is fine!! My spell-checker thinks it's okay too...!!\n; )\n\nIf you offered me an 'octopie' - I'd think it was covered in pastry!!\n\nOh, and it's 'spelt' OR 'spelled'!! BOTH are correct...!\n; )
450313
10 percoset should do the trick\nThis is a joke....just in case you are dumb enough to try it.lol
830557
Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
593300
8 squirts.\n8 works.\nmaybe even 9.