the clue is organization or association(abbr.). it has 4 letters and i believe that these are some of the letters- \ni-n-blank-t but im not sure...can you help me? i really need figure this out!!
136330
inst. for institution. That sounds right.
522588
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)
418677
I don't understand your question.  Do you mean using the associative law?\n\nabc*de = \ncba*de =\nbca*de =\nacb*de =\ncab*de = \nbac*de \nNow all of these are multiplied by the two ways to postiion de\n\nThat's 3! = 6\ntimes  2! = 2\n\nfor a total of 12.  \n\nFor a 5th grader I'm sure that you're just supposed to list all 12.  For convenience, note that I've put each letter in the middle and switched the other two.  For neatness, I'd recommende just putting another column to the right of each of these 6 with de and ed .\nWhen doing counting like this, called combinotorics, it's helpful to arrange things neatly.
196468
Number of ways to arrange ABCDEF = 6! since all of the letters are distinct.\n\nNumber of ways to arrange TOYOTA = 6!/(2!2!) since there are two T's and two O's.
303963
Good Luck mandylaz----I have not a clue!! I am not very good when it comes to throwing numbers and letters together!! :) Sorry!! Hope you find your answer!!
549429
[i]  = I\n[d]s = does (without the s it would be do)\n[t]  = to or the\n[o] = of\n\nAnyway I think all the [ ] are suppose to be abbreviations and these are the most likely.◘
96829
For the first question, that means you want to know about committees with exactly 3 women and committees with exactly 4 women. Since these are independent cases, we consider each separately and add the result. \nFor 3 women, we need to know how many unordered groups of 3 women we can choose from 4 women, so we have 4[Choose]3. The men have 2 people on this committee, so they contribute 4[Choose]2 to the product for a total of (4[Choose]3)*(4[Choose]2) committees with exactly 3 women on board. \nThere is only one way to grab an unordered set of 4 objects from a set of 4 objects, so we go right to the men. There are 4 ways the men can fill the last spot, so this is 1*4=4 different committees with 4 women on board.\nThus the total number of committees with a majority of women on board is (4[Choose]3)*(4[Choose]2) + 4.\nThe second part is very similar to the first.
671851
that cannot be answered because there is no solution, without the solution A, B, and C could be anything\n\njcr_0830_2005, how do you know the answer is 12? not only that there must be 3 different answers, in algerbra, the task is to figure out what # the letter should be
304173
P, Q, R, S, T, U, V, W, X, Y, Z!!! Did I get it right???
234093
I think it should have been C for Charlie, but if A is Charlie then here are the events.\n\nCharlie chosen ist:\nABC ABD ABE ACD ACE ADE ACB ADB AEB ADC AEC AED\nCharlie chosen 2nd\nBAC BAD BAE CAD CAE DAE CAB DAB EAB DAC EAC EAD\nCharlie chosen 3rd\nBCA BDA BEA CDA CEA EDA CBA DBA EBA DCA ECA DEA\n\nTotal events 36. This is correct since the the number of distinct possibilities is 5x4x3 = 60\nAnd the no of ways he won't be chosen are 4x3x2=24\n60-24=36
66315
Seems like a pretty interesting riddle. I am so sorry that it didn't get the response it deserved. I think you should have placed it in the "Jokes and Riddles" sub-category under the category "Entertainment & Music". It would have garnered an immense response there.\n\nAnyways, let me try here-\n4 is golden,eh? I think here 4 somehow refers to "silence"\nNow, 5 is silence. Ummm.... what should be 5? I give up.\n\nGood Riddle. Try posting it in the "Jokes & Riddles" category.
52724
Hey asker, this is a great puzzle.  Could you provide the correct answer after a couple of days?  I am really curious.  Thanks.
439752
Alrighty! This one has been floating around for a very long time and this is the Third instance of the same riddle i've seen. I will cover it accordingly and supply the Answer so bare with me\n\nThe Riddle!\nCorrect Version....\nIn all instances of this riddle you notice that the title states "Find the Error" , Or "Find The Error For Me", or "Can you find The Error".  The Title is the KEY.. Notice how it just ask you to "Find The Error" it does NOT however state "What" kind of Error.. This is pretty much the meat and potatoes of the Trick Question..\n\nIncorrect Versions...\nNow there are some instances that actualy go as far as to say " Find the Error in The Alphabet". These titles are incorrect and defeat the purpose of the Riddle.. Fact of the matter is There IS NO ERROR in the Alphabet, or should NOT be.. But since the title was posted wrong, this confuses the Readers to the REAL problem and answer..\nSome have even "Intentionaly" distorted the alphabet by adding a "0" into the mix instead of the "O". Now this is tricky, yes, but once again. Incorrect, and defeats the purpose of "Misdirection"\n\nAlso another mistake people make is they "Repost" The riddle by hand INSTEAD of just copyiing and pasting it  the way they should. So it gets distorted as it moves on. This case is NO exception. The mistake made in this instance is with the sentence \n"find the error. it's impossible" <- this is the incorrect version\nit should be\n"find the error. its impossible"<-This is the Correct version \nNotice how in the "Correct Version" of the Riddle, I Intentionaly Neglected the Apostrophe between the "t" and "s" :) That is because THIS is the Error the reader is supposed to find. It's a grammatical Error NOT an Alphabetical.\n\nThe Answer!\nThe Error Is Not In The Alphabet\nNow since we know what the Real Version Is, let's cover the Answer.  As said.. the Title is a "Trick" Question.. when stated correctly, the Reader "Assumes" that the error WANTED is in the Alphabet , even though it was not Directly Stated, so you search the entire Alphabet for Anything that could be possibly seen as "Wrong" but this is intentional  Miss direction..\n\nThe Error SHOULD be in the actual question ABOVE the alphabet list .\n\n"find the error. its impossible" \n\nNotice how in "Its impossible", there is no apostrophe between the "t" and "s".. At least when this riddle is posted "Correctly", there should be "NO Apostrophe"..\nThe absences of the apostrophe is the "Error" of which the reader is supposed to find..\nConclusion..\nWhen you repost this riddle be sure you just have the title read "Find The Error" and try not to convert it in any way.\nAlso never try to retype the riddle by hand unless you KNOW exactly how it's stated. People already tend to make grammatical errors when typing as is and others tend to make NO mistakes whatsoever , either way , this this distorts the Riddle...
285125
It is a major puzzle, no one is quite sure. See the attached article for more details...
140
Permutation\nSuppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.\n\nBecause order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:\n\n      CA   CT   AC   AT   TC   TA\n\nCombination\nWhen we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:\n\n      CA   CT   AT
366480
look very carefully at the clue.  the misspell is there for a reasonnn
516885
C is at the end, so its just the other 8 positions to play with.\n\nCount how many permutations of those 8 there are.\n\nThen count how many there are where A and B stand side by side.\n\nThen subtract those out.\n\nEdit:  Poster below me is way off.  without any restrictions, 9 men could be arranged in over 360000 ways.   Not 81.\n\nBasically, you can take 'C' out of it.  There are 8 spots left, and 8 men.  Spot 1 can hold any of the 8.  spot 2 any of the remaining 7 and so on, giving 8X7X6X5X4X3X2 which is 8! possibilities.\n\nThat's 40320.\n\nOf those 40320, some involve A and B being side by side.  How many?  Well, they could be in slot 1&2, 2& 3 .... up to 7&8.  And they could be A-B or B-A.  That's 14 possibles.  For each of those 14, the other 6 spots could be filled 6! ways.\n\nSo there are 14 X 6! = 14 X 720 = 10800 combos with C in the 9 spot and A & B side by side.\n\nTherefor, of the 40320 with C in the 9 spot, 40320-10800 = 30240 possible arrangements have C at the end, but A and B NOT side by side.