How many different ways can you paint a cube using only one color on a face if you have red paint and blue paint? Two cubes are alike if they can be rotated so one matches the other exactly, color for color. One cube could be painted all red and another all blue. There are others. For example a cube could be painted with four red faces and two blue ones. This one can be done more than one way as can some of the others. Another cube could be painted with four blue faces and two red ones. There are more. Give the grand total number of ways the cube could be painted
208898
This is very tricky. Start off with the basics. Use Red and Green, 2 cubes and count sides of color.\n\ncube1: gggggg\ncube2: rrrrrr\n\nrrrrrr - gggggg\nrrrrrr - gggggr\nrrrrrr - ggggrr\n.\n.\nrrrrrg - gggggg\nrrrrrg - gggggr\n\nTake out all the similiar ones to find the count.\n\nInteresting question. Probably easiest to run a recursive computer simulation to find out.\n\nTotal for 1 cube = 10\ngggggg - 1 way\nrggggg - 1 way\nrrgggg - 2 ways\nrrrggg - 2 ways\nrrrrgg - 2 ways\nrrrrrg - 1 way\nrrrrrr - 1 way
79099
wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
200938
To find the surface area of a cube without measuring a face, you would:\n1) Multiply: length*width\n2) Multiply: length*height\n3) Multiply: width*height\n4) Then you add those three answers up \n5) After you add them, multiply by 2 because a cube has 2 of each face.\n\nHope this helps!!
390779
It depends if one person can be in a room or if all 4 people can be in the same room.  If only one person can be in a room at any given time, then it would be \n4*3*2*1=24 different ways\n\nIf everyone can go in the same room, then there are 4^4=256 different ways.
159709
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
63648
You must think about the 3d visualization of 4d objects as partials projections or perspectives (as well as when you represent a cube drawing it on a paper sheet, you only describe some of its properties by analogy): so you cannot "see" 4d objects: you can only describe them.\n \n\nSee this link for a rough idea of an hypercube (there would be much more to say about it ...)\nhttp://answers.yahoo.com/question/index;_ylt=Apvwu5u3VyjMZfW_mDnnJUnsy6IX?qid=1006012001214\n\nSee this other one (expecially the links in all answers) for the so-called "Klein bottle"\n\nhttp://answers.yahoo.com/question/index;_ylt=As6vB06qMWLWdUaS3dhCfW_sy6IX?qid=1006013107146\n\n\nfour dimensions is interesting in maths because (they say: I'm starting now to study it) that there are many different "regular solids" which you can hypotize counting sides, vertexes and faces. From 5th dimension on, there is only one kind of regular solid, so it is not considered an interesting field of study.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
368105
The total number of cubes will be:\nn * (n + 3) * (n + 9) cubes.\n\nTo get the number of cubes inside, subtract 2 from each dimension:\nInside cubes = (n - 2) * (n + 1) * (n + 7)\n\nNow expand these out...\nn * (n + 3) * (n + 9)\n= n (n² + 12n + 27)\n= n^3 + 12n² + 27n\n\nInside cubes:\n(n - 2) * (n + 1) * (n + 7)\n= (n-2) * (n² + 8n + 7)\n= n^3 + 8n² + 7n - 2n² - 16n - 14\nGroup like terms:\n= n^3 + 6n² - 9n - 14\n\nRemember that the inside cubes are exactly half of the total cubes, so double it:\n= 2n^3 + 12n² - 18n - 28\n\nNow equate this to the total cubes: \nn^3 + 12n² + 27n = 2n^3 + 12n² - 18n - 28\n\nPull everything to one side (so subtract n^3 + 12n² + 27n):\nn^3 - 45n - 28 = 0\n\nPossible answers for n will be the factors of 28:\n±1, ±2, ±4, ±7, ±14, ±28\n\nBy inspection, 7 works:\n7^3 = 343\n-45 * 7 = -315\n343 - 315 - 28 = 0\n\nSo the answer is:\n7 x 10 x 16 = 1120 blocks\n\n(Note: inside you have 5 x 8 x 14 = 560 blocks)
406606
The first letter can be any one of the 5 letters\nThe second one can be any of the 4 remaining letters\nand so on\n\nSo the answer is 5 x 4 x 3 x 2 x 1 = 120 different ways\n\nIf two of the letters are the same (eg green) then divide by (2 x 1) to give 60\n\nIf three of the letters are the same (eg eerie) then divide by (3 x 2 x 1) to give 20\n\nIf there are two pairs of letters (eg yuppy) then divide by (2 x 1) x (2 x 1) to give 30
455663
Total number of balls: 7\nLet's say the player draws each ball with each hand.\nThere can be 2 situations:\n1)Player draws 1 red ball with left hand and 1 black ball with right hand: (2/7 * 5/6)\n2)Player draws 1 red ball with right hand and 1 black ball with left hand: (5/7 * 2/6)\n\nTherefore, probability of drawing exactly 1 red ball is (2/7 * 5/6) + (5/7 * 2/6)= 10/21
63215
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
356939
Really Good Question.\n\nThe answer is 3 socks.\n\nExplanation : Taking the case of least probability, suppose the first sock is brown and the second is grey, then the third has to be either brown or grey as there are only two colour sock. So the purpose is solved by drawing 3 socks.
624127
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..
511879
OK, this question doesn't make sense, for two reasons:\n\n1.) Colors, by definition, are at different wavelengths. If the beam has 2 color components, each component would have it's own wavelength.\n\n2.) Your question at the end isn't a proper sentence, and is leaving a necessary part out.
84461
just mix 'em up  , if you give me the name of 'all ' the colours , i will tell you how to make them\n\nI guess you mean primary colours (= base colours)\nthese are yellow red and blue\n\nyellow and red will make orange\nyellow and blue will make green\nred and blue will make purple\nyellow , red and blue will make brown\nmix them up in different proportions to get all the other colours
497018
This is the old four color problem of topology.  Mathemeticians showed years ago that no more than four colors can satisfy a map where each country is a single unit.
562445
you use both halves of your brain. you use different halves during different activities. in artists the right brain is the stronger side. when you are doing a math problem you use a different 1/2 of the brain than when you are painting. they both do the same thing in everyone. it is just in some people are stronger in one half. people are not just either logical or creative. that is not how it works. you are both but you may have a strength.\n\nnsrunner16@yahoo.com