question! arrange is sequence.?
79607
First, I would turn off the faucet so I won't waste water. Second, I would look out the window to see who's knocking on the door then let the person in if it's someone I know. I would leave the laundry outside because it's going to dry after it stops raining. Last, I would wait for the answering machine to pick up the message and then I would call the person back if I know that person.
266405
The order of operations gets messed up.  You have to add x+y before you square it.\n\n PEMDAS \n  (You might remember this as "Please excuse my dear Aunt Sally.")\n\nOrder of Operations\n\nParentheses \nExponents \nMultiplication and Division \nAddition and Subtraction
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A\nB\nA\nD\nC\n\nIn that order.
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Combinations are basically what you do when order doesn't matter.For example, electing any 12 students to a group of 3. It would be 12 C 3.\n\n\nPermutations are what u do when order DOES mattter. For example, when your arranging 6 different books. The answer to it would be 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways.\n\nFactorials are basically. The number you have multiplied by everything less than that number.For instance, if you have 3. The factorial notation would be 3 x 2 x 1 = 6
18305
They are in alphabetical order.\n\nE,f,f,o,s,s,t,t,z
483473
there might be a - or a . in the sequence
574164
Eight\nFour\nNine\nOne\nSeven\nSix\nThree\nTwo\nZero\n\nThe numbers are in alphabetical order. But why is 5 not there?
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It's called the Fibonacci sequence, but there's actually a couple more terms at the front: 0,1,1,2,3,5,8,13...  You take the last 2 terms and add them together to get the next term.  So 8+13 = 21; 21+13 = 34, etc.\n\nHere is a good start if you're interested in learning more about it:\n\nhttp://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/
517585
Starting with 0, the sequence is the square of that number:\n\nf(n) = n^2. \n\nf(0) = 0 (0^2)\nf(1) = 1 (1^2)\nf(2) = 4 (2^2)\nf(3) = 9 (3^2)\nf(4) = 16 (4^2)\nf(5) = 25 (5^2)\netc.\n\nNote, it isn't clear if your indexes are going 0,1,2,3,4,... or are going in descending order 0,-1,-2,-3,-4... the resulting sequence would be the same.\n\nf(0) = 0 (0^2)\nf(-1) = 1 (-1^2)\nf(-2) = 4 (-2^2)\nf(-3) = 9 (-3^2)\nf(-4) = 16 (-4^2)\nf(-5) = 25 (-5^2)\netc.
410291
http://en.wikipedia.org/wiki/Permutation
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Nine (9) - before 10 and after 11\nOne (1) - before 2 and after 0\nOne (1) - before 2 and after 0
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If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
1103012
35\n\nThe sequence is every even-indexed Fibbonacci number plus 1.  So from the original sequence whose value is the sum of the two previous numbers:\n\n0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...\n\nYou take every second number and add 1:\n\n2, 3, 6, 14, 35, 90, 234, ...
286004
What is the relationship between the items in the sequence? If you are adding the same thing over and over, it is arithmetic (2,4,6,8,10). If you are multiplying by the same thing over and over, it is geometric (2,4,8,16,32).
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7 will be after five\nand eight will be after six.\n\nReason: The upper numbers in 'numerals' contains straight lines, the lower one contains curves.\nWell...
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So is this sequence based purely on a mathematical function?\n\nOkay, so there's a 52 in there, could this sequence somehow relate to the calendar?\n\nWhat sort of scientist? I'm a computer scientist and nothing's glaringly obvious about those numbers.\n\n** GOT IT *** 68! Er or Erbium, the sequence spells out Brainteaser based on the atomic weights of B, Ra, In, Te, As, and Er.
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1) 21\n2) 52\n3) 40\n4) 80\n5) 104\nIn each pattern you are adding a number to the one before it.\nEx: 1 + 0 = 1, 1 + 1= 2, 2 + 1 = 3, 3 + 2 =5, 5 + 3 = 8, etc.