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![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jul 2008 21:34:10 IST
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Twenty-seven identical white cubes are assembled into a single cube, the outside of which is painted black. The cube is then disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man (who cannot feel the paint) reassembles the pieces into a cube. What is the probability that the outside of this cube is completely black?
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11 Jul 2008 22:06:57 IST
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plz try at least:(
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11 Jul 2008 22:09:54 IST
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0 i think. I don't know if any of the squares will get black completely.
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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11 Jul 2008 22:24:28 IST
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um...how can you arrange 27 cubes into a bigger cube? i can't seem to be able to arrange them!
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Alcohol and Calculus -- DON'T MIX -- Please don't Drink and Derive!!!
A committee is a group of people who individually can do nothing but together can decide that nothing can be done.
MUNNA BHAI: Circuit, bole toh yeh Ford kya hai?
CIRCUIT: Bhai, gaadi hai.
MUNNA BHAI: Toh phir, yeh Oxford kya hai?
CIRCUIT: Bole toh, simple hai bhai, Ox mane Bael, Ford mane gaadi. Oxford bole toh Baelgaadi. |
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11 Jul 2008 22:43:47 IST
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@rudra.....
it cant b 0 obviously.....
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11 Jul 2008 22:46:50 IST
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You are asking that the probability the cube has all its 6 faces black?
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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11 Jul 2008 22:53:54 IST
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hmmm...
there'll b
12 smaller cubes vth 2 black faces
8 with 3black faces
6 vth 1 black face
1 vth no face black
we'll consider each category separately......
now one 2 face colored cube can occupy the position of anothr 2 face colored cube....nd so on......each can b fitted in2 d place of anothr of itz kind nd not newer else......
so we find the number of ways.....fer each category.....nd divide by total no.of ways.......nd there u have it!
this is basic idea!
vll post complete solution l8r......am i r8 little_genius??
PS: dont 4get the along vth position of cube color of d side mus also b taken in2 account!
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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11 Jul 2008 22:57:59 IST
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well i have got an answer...
but i dunno if it is correct..
it is .. 1/(24*27*27*27*27)
i knw it is really absurd... but if it is correct.. i will post the solution.
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11 Jul 2008 23:12:30 IST
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well i dont know the answer myself.....just thought to post it so that all cud try .......me too trying it....
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11 Jul 2008 23:53:42 IST
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awryt....itz a try....
k m assuming dat he is buildin d cube from bottom up....row by row......
for bottom row---he needs4 vth 3face colored......4 vth 2 face colored nd 1 vth 1 face colored.......
he picks em out in
8C4 * 12C4* 1C1 ............nd aftr selection....the 3color cubes can b placed on black face in 3/6 ways......nd rotated in 1/4 ways.......------>so in 1/8 ways
2 color cubes can b placed in 2/6 ways nd rotated in 1/4 ways.......---->>1/12
1color cube can b placed in 1way and rotated in 1/4----->>1/4.......
proceed lyk dis fer othr rows!
m still not sure
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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