In that case we are done for the day.

Aarav, thanks for the admiration. But kindly don't reduce me to an IITian!

Math Honours? That explains. From where? I was doing my M.Sc Math from I.I.T Bombay.

That is correct Aarav. Did you study Pure Mathematics?

selective_sloth , y shld one count identity map as only 1 , it is possible to have different functions for (1,1) (2,2) (3,3) (4,4)

?

I am referring to the binary tree of possible score-lines. The root of this tree is 2 - 0 and the leaf nodes are all 5 - 4. At each node there are at most two branches. Hence, binary tree.

Part A -- Answer is (5)

If f:X -> X where X = {1,2,3,4} such that f.f (x) = x, then f is a bijection. There are 24 bijections from X to X. Of these only 10 are such that f.f (x) = x for all x in X.

In giving these maps explicitly I will use a cyclic notation for convenience.

eg, (1 2) stands for the map 1 -> 2, 2 -> 1, 3 -> 3, 4 -> 4

Note that in this cyclic notation 3 and 4 are not mentioned which means that they are mapped to themselves i.e. they are fixed.

The maps are :

(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)

and also

(1 2) (3 4)

(1 3) (2 4)

(1 4) (2 3)

Note that (1 2) (3 4) means 1 -> 2, 2 -> 1, 3 -> 4, 4 -> 3

And lastly the identity map which gives us a total of 10 maps satisfying the given condition.


Part B -- Answer is (3)

If one draws a binary tree to represent the problem, one is required to only count the number of leaf nodes of the tree. They turned out to be 14 in number.

So far.. all the answers are wrong. Somebody has factored n^4 + n^2 + 1 which was a previous problem in this forum in place of n^4 + n^2 + n which is required here.

Why is the ratio of the two diagonals 2:1?

And of course.. there is that thrill that you sometimes get when you cheat..!