What is the probability of three cubes chosen out of a cube of n * n * n order, such that they lie on a diagonal??
This is modification of richa's Q, and this req 3-d view.....
9for a cube
12 facediag
4 body diag
what other diagnols are u talking abt ??
1i think this will help(may be)sorry if not
In ncert chemistry text book... in solid state the atoms will be view in diagonal manner in case of hcp packing..((2 coloured pages with the arrangement of balls and type of close packing)....Visualise like that fromm each vertex of the cube...
1no richa... any 3 collinear cubes need not be diagonally arranged.....
1Clarification needed.
What is meant by "probability of 3 cubes"-------- does this mean cube of order (1*1*1)?
Morover what is meant by lying on the diagonal? For example, if a vertex of a cube lies on a diagonal, do we consider the cube to lie on the diagonal?
If the question means (1*1*1) cubes, and if lying means that diagonal passes through the cubes, I think there are 16n-33 such cubes.
I might be terribly wrong----- because I tried to count them. :-)
1sri do u mean any three collinear kind of cubes r considered to be on diagonals
1u leave 2d diagonals.... in 3d diagonals... those 4 are major ones... i also mean the smaller ones ,,,just imagine... i cudnot draw!!
1by diagonal if u mean the one connecting (1,0,0) and (0,1,1) {in a cube having sides (1,0,0),(0,1,0),(0,0,1),(0,1,1),(1,1,0),(1,1,1)}
then we have 4 diagonals...,for each diagonal u have n cubes, and the total number of cubes is n3...sample space is
n3C3.....and event space is 4*nC3
(as we have 4 such diagonals)
so P(E)={4*nC3/n3C3}
if by diagonals u also mean the normal 2-d diagonals on the faces of the whole cube, den the answer wud be different...
cheers!!!
1i cudnt explain wit words....i dont hav ssome software to draw those things.... wait i think nishant sir has one such software.. i will ask him to help
1it can be both....
see the cube has 4 major diagonals and there are other diagonals ... i mean diagonally arranged cubes it is not only of major diagonals... so all the cubes will be a part of one or other diagonal... how to count it... thats my problem
1can u explain this...even i dont know the answer... i want the mehtod...
No. of cubes lying on a diagonal = 4n-(n-1) = 3n+1
106total no. of cubes = n3
Any selection of 3 = n3C3
No. of cubes lying on a diagonal = 4n-(n-1) = 3n+1
Selection of any three cubes from this = 3n+1C3
So, prob = 3n+1C3/ n3C3
im not really sure
ive considered the cubes we select lie on the body diagonal only.