How do you get all of the same colour on the same side of a rubics cube. if there is a patteren.
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Check this out. This is the best site I've found so far. It helped me solve the Rubik's Cube in just four days: http://lar5.com/oldcube/index.html
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It's called ASCI art. There's a software that can produce them using software called bg_asci
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in a cube, there are 6 sides, in each side, there is only one number, thus called a one number cube.\n\nThe order is 1-6. number 3 comes out only once. Thus, the chance of getting a 3, is only one out of 6. making it 1/6 in math language.
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Yes its the Rubik's cube. I remember when it was popular. I hated it. There was even a TV special on 101 ways to destroy it. Maybe that is why I am a wee bit off.
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Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
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Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
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you can buy rainbow pasta in almost any supermarket (red, green and the regular yellowish colour).  i think you'd have to dye it if you wanted other colours.  there are also fun shapes like bowties and wagon wheels.  i know wagon wheels come in the tri-coloured variety because we just had some for our evening meal tonight.
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It cannot end in the middle.\n\nLet's look at the cube.  \nIt's divided 3x3 to give you 27 subcubes.\nThe middle cube is therefore only one deep from any side.\nThere are eight subcubes on each corner of the cube.\nThere are six subcubes in the middle of each face of the cube.\nWe'll put all these subcubes into a group.\nCall it group A. 8+6=14 subcubes.\n\nAnd there are all the other cubes.\nThere are 27 subcubes.  14 are in group A.\nSo there are 27-14=13 of these.  \nWe'll group them together and call it group B.  13 subcubes.\n\nSome things become clear at this point.\nThe mouse could progress always ahead. So...\nEach time the mouse eats one of the group A subcubes, it must go to one of the group B subcubes.\n\nEach time it eats one of the group B subcubes, it must go to one of the group A subcubes.\n\nEven if it goes sideways, to a subcube in the same group (that's only possible for group B subcubes because they are the only subcubes where members of the group can be adjacent), what happens?\n\nThe number of subcubes in group B will reduce by two instead of one.\nThe number of subcubes in group A will remain the same as before.\n\nHowever the mouse moves and eats, there will always be more group A subcubes than there are group B subcubes.\n\nThe middle subcube belongs to group B.\nIt's one of the group of 13 subcubes.\n\nWhichever group the mouse eats from on the first day, he will always have one of group A subcubes left.\nBecause there are always more group A subcubes than group B subcubes.\n\nI'll repeat all that.\n27 subcubes, one per day = 27 days. \n13 subcubes in group B.\n14 subcubes in group A.\n\nIt is impossible for the mouse to have eaten the last group B subcube and there be no group A subcubes left.\nBecause there are always more group A subcubes than there are group B subcubes.\n\nSo the last subcube the mouse eats will always be one of group A.\nA corner subcube or a subcube that's in the middle of one of the faces.\nNot the subcube in the middle.\nQED, I think.
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1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
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Try Cube: http://www.cubeengine.com
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You may not realize it but in one qhestion you have asked for knowledge that is gained taking a university level physics course.\n\nSince I am a lousy typist, I will give you some clues that should allow you to visit Google and gather the answers you are looking for.\n\nThere are twi type if color mixing.  The additive method, used by painters, where you mix yellow and blue to get green.  \n\nThe subtractive method, used in optivs and video technology, where colors are filtered out as opposed to added in.  In this method the primary color are differet.\n\nIf you want a much more thorough explanation, you may want to examine the study of "Chromatics".  At look at how rainbows are created might help as well.\n\n\nBest of luck,\nJeffrey
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I think you are referring to magic numbers making up a magic square wherein you arrange such numbers in a grid and get the same sum for all the numbers in the same row, same column and diagonal.\n\nFor example, take the numbers 1 -9 and arrange them in a 3 x 3 grid (3 rows and 3 columns).  Set the 5 in the middle and arrange the rest around it so as to get a total of 15.  You should get something like this arrangement.\n\n     6     1     8     =     15\n     7     5     3     =     15\n     2     9     4     =     15\n\n     =     =     =      two diagonals each add up to 15 as well\n    15    15   15
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use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
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Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.  \n\nFirst, we can prove the 5 has to go in the middle:\n\nLet the numbers in the top row from left to right be a, b, and c.  Let the numbers in the middle row from left to right be d, e, and f.  (Note e is the middle box.)  Let the numbers in the bottom row from left to right be g, h, and i.     \n\nWe know a + b + c = 15 and g + h + i = 15\n\nWe also know a + e + i = 15, b + e + h = 15, and c + e + g =15.  Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45.  We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45.  Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15.  Therefore e (the middle box) must equal 5.   \n\nKnowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15.  So, the 9 has to occupy a side box.  Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9.  This forces the placement of the 1 in box h.   \n\nWith the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1.  So, we'll put the 8 in box g and the 6 in box i.  (We still haven't done anything to create a second square if rotations and mirror images are discounted.)  That forces the placement of the final four numbers.
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Make a Dodecahedron. That's what I did.
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8. One for each corner of the cube. Think about how many corners are on a die or a box. Each of these is a point.
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If there are Colours Join the Colors same to the same... Like I mean join the same colors... together...\n\nok.. i actually have no idea..\n\nMaybe it would help..