Doubt from permutations and combination


A scanning code consists of a 7 \times 7 grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of 49 squares. A scanning code is called \textit{symmetric} if its look does not change when the entire square is rotated by a multiple of 90 ^{\circ} counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}


Let [r,s] denote the least common multiple of positive integers r and s. Find the number of ordered triples (a,b,c) of positive integers for which [a,b] = 1000, [b,c] = 2000, and [c,a] = 2000.


A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?


Is it option b (1022)

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Yes sir ! , Show solution





A small hint for imagining the polyhedron :
Consider this. We know a cube has 6 faces, 8 vertices and 12 edges
Similarly for this polyhedron we have 6 octagons along the six faces, 8 hexagons corresponding to 8 vertices and 12 edges corresponding to 12 edges.
Now try. If you are unable to solve further do tell

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What answer are u getting bro
I'M getting different from given


Here the best I can think of is that at least one of the numbers is either 2000.
It is not sort of a proper method. But I am getting 34

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Okay I guess imagining the structure would be complex and will not help us in getting the answer.
So here's the way I did: (there's a high chance I am wrong :sweat_smile::sweat_smile::sweat_smile:)
First calculate the total no. Of points:
6 octagons = 6×8 = 48
8 hexagons = 8×6 = 48
12 squares = 12×4 = 48
So total 48×3 points would be there if each vertex had been separate
Note that each vertex is shared by a hexagon, octagon and a square.
So we effectively have only 48 points
Now for inner diagonal, each point will have access to 36-1 points as we need to exclude the vertices of square, hexagon, octagon and -1 for the point itself.
So now we get 35 for one point for the next we will have 34, 33 and so on.
So total 35+34+33+32.. +1

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Answer is 840 @Mayank_2019_1 bro

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For which one ?
Please clarify @Tushar_2019


Okay got it ... Didn't do that properly.
What I missed was that when each point is connected to 35 different points then we don't get the points that are there in the plane of square, hexagon, octagon. and then we can't subtract 1 from that and say that the next point will have 34 connections.
so the solution:
As explained above the polyhedron will have 48 vertices. By the same logic we will have 483/2 = 72 edges.
We will first count all the connections so we will have 48+47+46+45+...+1 = 1128 connections
Now we need to exclude the ones which are in the plane of a point. So first we exclude all the edges we have 72 edges.
For 1 square we have 2 diagonals so 12
For 1 hexagon we will have 3+3+2+1 = 9 diagonals so 89=72
For 1 octagon we will have 5+5+4+3+2+1=20 diagonals so 6
So total 72+24+72+120
On subtracting the above sum form 1128 we get the desired result,

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What is the answer of the question posted by you in post #2 ?
Is my answer correct?


@Mayank_2019_1 your answer is absolutely correct for #3 and answer for #2 is 70


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