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rajeshmarndi
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Homework Statement
A= {a1,...am}, B= {b1,...an}. If f: A→B is a function, then f(a1) can take anyone of the n values b1,...bn. Similarly f(a2). Then there are nm such function. I understand this part.
So in my book, using this principle,
nC0 + nC1 + ... + nCn = 2n is proved.
It has taken a set A= {a1,...an}. For each subset B of A, there is a function fB : A→ {0,1} given by fB(a1) = 1 if a1 ∈ B and fB(a1) = 0 otherwise. Conversely, for every f: A→ {0,1}, B = {a1 : fB(a1) =1 } ⊂ A anf f = fB. Thus there is a one-to-one correspondense between the subsets of A and the functions f: A→ {0,1}. From the above formula, hence we have
nC0 + nC1 + ... + nCn = 2n