Symmetry - II

I just love solving things by symmetry. I feel as the variables increase in any problems, symmetry can play bigger and bigger role. Another CAT type question:
The average value of Mod (a - b) + Mod (c - d) + Mod (e - f) for all possible permutations a, b, c , d, e, f of 1, 3, 5, 7, 9, 11 is :
My first reaction when I saw this problem, I will not attempt this unless I have no other question left. But once again I thought so many variables - must be some easy way out. I feel these problems can be solved in steps making them simpler with each step. I mean what is the use of Mod here - when values of a or b and c or d and e or f can be interchanged. As in once we choose the pairs (a,b), (c,d) and (e,f) such that values of (a and b), (c and d) and (e and f) can be interchanged - we can form 2*2*2 combination of the expression with interchanging values of and b or c and d or e and f. Also, because of Mod function outcome of the expression for all 8 ways would be the same. Out of these 8 case there surely will be one and only one case where a>b, c>d and e>f. Hence, if we just choose this one case and ignore 7 other cases as they do not add or subtract anything to symmetry as outcome of all the 8 are same and when we average it, it is equivalent to multiply for one particular a - b + c - d + e - f by 8 and then dividing by 8. Hence the expression can be reduced as (a-b) + (c-d) + (e-f) where a>b, c>d and e>f.
Hence, now we have to find Average (a-b + c-d + e-f) which can further be simplified to Average (a-b) + Average (c-d) and Average (e-f).
Now the beautiful symmetry: Average (a-b) must be equal to Average (c-d) must be equal to Average (e-f)
So the expression reduces to 3* Average (a-b)
Now the problem has just reduced to plug the numbers and get the average.
If a=11, b can be 1,3,5,7 and 9 and (a-b) = (10,8,6,4 and 2)
If a=9, b can be 1,3,5 and 7 and (a-b) = (8,6,4 and 2)
If a=7, b can be 1,3 and 5 and (a-b) = (6,4 and 2)
If a=5, b can be 1 and 3 and (a-b) = (4 and 2)
If a=3, b can be 1 and (a-b) = 2
Hence average = 3*((10,8,6,4 and 2)+(8,6,4 and 2)+(6,4 and 2)+(4 and 2)+2))/(5+4+3+2+1)
= 3*70/15 = 14 and ofcourse which is the correct answer!
This problem should take two minutes but solvingthese kind of problems do help in making the concepts clear in your mind. This problem appeared difficult in the beginning, but I always believe that if its unsolvable in two minutes, it will not be in the CAT paper. So there is an easy way out always - practice and deriving solutions on your own with your own methods and technique will surely help in finding that way.
Just to post the alternate solution on some CAT related website which I found was a bit complex : By symmetry the average of a - b is independent of choices in which they are chosen (provided they are unequal). Suppose a = k. Then the average of k - b is (k-1 + k-3 + ... + 2 + 2 + 4 + ... + 11-k)/5 = 2*(k^2 -7k + 21)/5. So average of a - b is (1/6)*2*sigma (k^2 -7k + 21)/5. Hence required average is 2*sigma(k^2 -7k + 21)/10 = 2*(91 - 147 + 126)/10 = 14.

Symmetry

"Symmetry generally conveys two primary meanings. The first is an imprecise sense of harmonious or aesthetically-pleasing proportionality and balance; such that it reflects beauty or perfection. The second meaning is a precise and well-defined concept of balance or "patterned self-similarity" that can be demonstrated or proved according to the rules of a formal system: by geometry, through physics or otherwise" (Source: Wikipedia)

I have always been a great fan of symmetry. Not only it pleased my aesthetic sense but always made me curious to find out more and more symmetrical objects around me. I am sure symmetry would have in store answers to a lot of our unanswered questions. Certainly it can make life easy in CAT mathematics - dont get surprised - trust me it can !!!

Lot of times you would see a question like how many times a particular digit would come say in first n natural numbers. I mean I cant think of using symmetry more beautifully in mathematics than solving this problem.

For example let us say the question was how many times digit 6 would come in first 986 natural numbers. Most of you would be able to solve this problem with the following approach:

In first 100 numbers, 6 would appear in 10 places at Units place, 10 places at tens place, so every 100 there would be 20 times 6 so till 1000, there would be 20*10 sixes in units and tens places. Add another 100 for hundres place between 600 and 699. Hence 300 times till 1000 and subtract 1 six becuase of 996 not being counted in first 986 natural numbers, and the answer is 299. Answer is correct but it does take time.

Now, try solving this problem using symmetry. Assume writing all numbers from 0 to 999 in three digit format i.e. 000, 001, 002, 003, 004...... 996, 997, 998, 999. In this way there are 1000 numbers each with three digits. Hence total number of digits are 1000*3 = 3000. Now the symmetry comes into action. Each of the 10 digits from 0 to 9 must repeat equal number of times, hence digit 6 should come 3000/10 i.e. 300 times and subtract 1 six because of 996 not being counted!! 299!!!!!!!!! voila!! was not it quickand smart!! Symmetry (and Sonu!!) doing the magic.

I am sure no coaching institutes would teach this (untill they read my blog!!). But, again the idea I want to convey is to start inventing solutions to the problems. That is the best way you will remember it and may also help in boosting your confidence.

Chess Board

I am not writing after a long time but posting after a long time. My last few write-ups did not see the daylight as they were not approved by the censor board as the content was highly controversial as well as personal for few people around me.

Someone asked me some CAT level mathematics questions. Suddenly, I realized CAT 2009 is coming and people must be working real hard to crack the claimed to be the toughest examination in the world to crack. Cat screws them all...

Coming back to the question, "Two squares are chosen at random on a chessboard, what is the probability that they have a side in common?"

Pretty simple for most of you - Most of you who would be taking CAT seriously will be able to solve this not so difficult problem. I am sure the approach would be different. CAT as I know is all about solving problems efficiently. I did ask this question to few of my friends who all gave me the correct answer and most of them approached the problem in the following way:

If the first square is any of the corner ones, then the second square can be chosen in 2 ways, if the first square is any of the remaining 24 squares on the sides of the chessboard, the second square can be chosen in 3 ways, and if the first square is any of the centre 36 squares, second square can be chosen in 4 ways i.e. (4*2 + 24*3 + 36*4) =224, and since all combinations are repeated twice, total number of combinations of chosing squares will be half of it. And Denominator of the required probability will be 64C2, Hence the answer would be (224/2)/64C2. Bang on the answer is correct. And at least 2 minutes in solving this.

But the approach efficient?

I mean there cant be any other choice for the denominator, but cant it be a little less clumsy way in calculating the numerator. I am sure it could be better than this. I think I would prefer it doing the following ways (order of preference in ascending order)

Just assume to put number 1 to 64, one each in the 64 boxes in the chessboard. Let the number be x for the chosen first box, then either x-8 or x-1 or x+1 or x+8 can be the possible second box which would satisfy that they have a side in common. Now to avoid double count we can reduce the number of boxes satisfying the second box to be x+1 and x+8 (because combination x-1 and x will be y and y+1 for some number y and x-8 and x will be z and z+8 for some number z). Now all the numbers from1 to 64 will have neighbours in the form of either x+1 or x+8, hence total combinations for the numerator should be 64*2 =128. But, since boxes 8, 16, 24.... 64 (right most column of the chessboard), do not have x+1 as their neighbour, hence we have to subtract 8, and also numbers 57 to 64 (bottom row of the chesboard, do not have x+8 as neighbour as chess board just have 64 boxes, hence we have to subtract another 8 and hence the possible combination for numerator should be 128-8-8 = 112 and ofcourse no prize for guessing the denominator. (I think it should not take you more than a minute)

Another way could be thinking (thanks to my friend Ramjo) that for each row, there can be 7 combination ((1,2), (2,3).... (7,8))satisfying the numerator and there are 8 rows so total is 8*7 = 56, similarly for each column there can be 7 combinations ((1,9), (9,17).... (49,57)) and there are 8 columns so total is 8*7 =56, hence total combinations for numerator would be 56*2 = 112. Another neat way of thinking !!(basically good for people who do not have algebraic bent of solving problems as suggested above but basically both are ultimately the same logical thought process)

One more way of approach could be for people who can visualize 3D geometry!!. Assume making a sphere out of the chessboard by joining rows 1 and 8 and colums 1 and 8. Now in this scenario, each square would have 4 adjacent squares which can fit to be the second square. Hence, total number of combinations for numerator would be 64*4/2 = 128. Now since rows 1 & 8 are not connected in the chessboard, subtract 8 combinations formed by rows 1 & 8 joined in the sphere. Similarly, subtract 8 combinations because of column 1 and 8 not being joined on the chessboard. Again the numerator would be 128-8-8 = 112. Few of us may have difficulty visualizing the sphere as I suggested out of the chessboard by joining extreme rows and columns !!!! But again it can make life easy and one sure way to approach !!

Similarly people may have n number of ways solving this problem. But again when time is essence, logical best and smart way can give you those extra seconds which can differentiate you from the rest !! Good luck ! My next post will be related to use of SYMMETRY - ofcourse that makes life beautiful.