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Q1: Give a bijective proof of the following identity E (%) = 2:1 using the following steps: (a) Use binomial paths represente

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Answer #1

We have   

now considering to be the number of paths on , only allowed the movements of path states , that is only allowed movements of going right by one unit and going upside by one unit , then the number of paths of total of n steps taking 2k number of steps towards right is where is the set of all the paths taking even number of steps to the right.

(b) Now defining to be the set of sequence of 0 and 1 , of length n, where number of 0 in the sequence is even

then the cardinality of the set R of the sequences of length n with only 0, 1 is , so the number of sequences with even number of zeroes in the sequences is just half of that of cardinality of R because of the symmetry of 0 and 1 , so

(c) Considering the bijection as defined as following

We will construct the sequence of 0 and 1 of length n by denoting 0 for each movement towards Right, and denoting 1 for each movement towards upside respectively, then is bijection from to

(d) The function is well defined as if two paths defined as above are different, then there is at least one position k, for which the one path has upside movement, and other will have right movement, and hence the corresponding sequences in will have 1 and zero respectively in that particular position , k-th position of the sequence, and for each path in we will find a corresponding sequence uniquely in , so the function is well defined.

Hence we have

Hence

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