A few days back, a student of ours asked me this question – in which cases is symmetry useful to us? Honestly, I don’t think I can create an exhaustive list of the topics where it could be useful. The first thing that comes to mind is of course, Geometry. Circles/equilateral triangles/squares/cubes are symmetrical figures. Symmetry helps us simplify questions which are based on these figures. We have also seen the uses of symmetry in dice throwing. In arrangements too, symmetry helped decrease our work substantially.

Today, let’s look at a puzzle where symmetry helps.

Question: A spider is sitting on one corner of a cube. It wants to get to the most distant corner but it can crawl only along the edges of the cube and cannot revisit a place where it has already been. In how many different ways can the spider go to the most distant corner?

(A) 6

(B) 12

(C) 18

(D) 24

(E) 30

Solution: The question is a puzzle type combinatorics questions. It seems like we will have to painstakingly calculate the various paths that the spider can take. But notice that the figure we have is a cube – a symmetrical figure. Let’s draw the figure to see what the question is asking.

Now, assume that the spider is at A. In that case, he has to go to F – the farthest vertex from A. Every vertex has only one vertex farthest to it. C, E and G are equidistant from A but they are in the same plane. F is further off than C, E and G. So it needs to go from A to F:

Step 1:

It can crawl only along the edges so from A, it can take three different paths – AB or AH or AD. As far as F is concerned, all the points D, H and B are similar.

Step 2:

Now, from each of these 3 points, the spider has two path options. If it is at D, it can crawl on DE or DC. The third path from D leads back to A but the spider is not allowed to revisit a place. So there are only two forward options for it – DE or DC.

Similarly, if it is at H, it can crawl on HE or on HG. If it is at B, it can crawl on BG or BC. So the total number of paths that we have found till now are 3*2.

Till now, we hope you did not face any problems.

Step 3:

Now comes the tricky part. The spider is at one of three vertices – E, C or G. Assume it came the AD – DE route and is now at E. There are multiple ways in which it can reach F. The obvious one is directly from E to F. But it can also go to F via H because it has not visited a number of other vertices (H, G, B, C)

There are three ways in which it can reach F now:

- directly E to F
- a three path EH – HG – GF
- a five path EH – HG – GB – BC – CF

This takes care of all the ways in which it can reach F from E.

Since we found 3 different paths from E, it is obvious that we will find 3 different paths from C and from G too. It is a symmetrical figure and hence we don’t need to calculate the number of paths from each point. In any case, we have 3 ways to reach F now.

So total different paths to reach the farthest vertex = 3*2*3 = 18

Answer (C)

Hope you see how symmetry helps us reduce our work substantially.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*