Calculate the number of arrangements of the letters in the word NUMBER?
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.
376218
Not sure how many you need, but here are the first few:\n2,3,5,7,11,13,17,19,23,29,31, 37,41,43,47,53,59,61,67, 71,73,79,83,89,97,101
681045
It helps to put them in order: 1, 1, 1, 2, 5, 5, 7, 8\nThen count in from the outsides to the middle, and it's halfway between 2 and 5, which is 3.5
33805
James should say 8!\nguard  s-i-x   = 3 letters\nguard t-w-e-l-v-e  = 6 letters\nso f-o-u-r-t-e-e-n = 8 letters
652572
This is a case of Circular Permutations where the formula is given by (n-1)! , taking into account clockwise and anti-clockwise arrangements as separate ones.\n\nSo your answer is 10 ! = 3628800 \n\nIf you are not taking into account clockwise and anti-clockwise ..then its 10 ! /2
100604
anything like, n.0... , n.1..., n.2..., n.3..., n.4... will be $n.\n\nanything like, n.5... , n.6..., n.7..., n.8..., n.9... will be $n+1.
405265
there are 17,576,000 possible arrangements. First, take the three letters, and find out all the possible combinations that there are for them. This answer turns out to be 17,576. Then take the numbers. There are 1,000 possible arrangements, so the number of possible arrangements for the license plate is 17,576,000.
768317
The letter o is a number
115468
Here's the pseudo code - code it in any language you want - \n1. Find length of word, say N.\n2. i = 0, flag = 0\n3. Repeat following step till i <= N/2\n4. If ith letter is not equal to (N-i)th letter, set flag to 1 and exit loop.\n5. If flag = 0, word is a palindrome.
326938
| | | | + - - - - -  = TEN\nthe question is really 4 lines + 5 lines = the words ten
1164476
The next number is 3. \nGoing by the number sequence and the number of letters per word in your question then that would be the next number. \nThe final number would be 3 as well.
612325
length = 9\nwidth = 9
803007
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.
889463
A pattern in numbers.  i.e. - 1,1,2,3,5,8,13,etc (add each number to previous number to get next number)
259152
words are from the GRE General Vocab list. numbers are the values for each letter. sorry there's no space in between them, but you can figure it out. \n\n\nhemoglobin851315712152914\nreciprocal185391618153112\napropos1161815161519000\nebullient522112129514200\nespouse519161521195000\nhirsute89181921205000\nimbroglio913218157129150\nmercurial135183211891120\npalpitate161121692012050\nplummet1612211313520000\nprofuse161815621195000\nrenovate185141522120500\nreprobate185161815212050\nsalacious191121391521190\nsubsume192121921135000\nsurcharge19211838118750\ntemperance205131651811435
934917
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.
370749
Arrangements: 9!/5!2!=1512. See Ron's answer for explanation.\n\nSelections: 12\nS and E are special, since there are more than one of each.\nSo we count each case seperately:\n4S        1\n3S0E      2\n3S1E      1\n2S2E      1\n2S1E      2\n2S0E      1\n1S2E      2\n1S1E      1\n1S0E      0\n0S2E      1\n0S1E      0\n0S0E      0\n\nAdd the numbers up, and we have 12 selections.