Prove of number of subsets of a set

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The discussion revolves around proving that the sum of binomial coefficients, nC0 + nC1 + ... + nCn, equals 2^n by establishing a correspondence between subsets of a set A and functions mapping A to {0,1}. It explains that for each subset B of A, a function fB can be defined, where fB(a1) = 1 if a1 is in B, and 0 otherwise. This creates a one-to-one relationship between subsets of A and functions from A to {0,1}. Participants express difficulty in understanding the proof's brevity and seek clarification on linking the subset concept to the function principle. The conversation emphasizes the need for clearer explanations to facilitate comprehension of the mathematical concepts involved.
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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

Homework Equations

The Attempt at a Solution

[/B]The prove given is so short for me to understand anything.
 
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rajeshmarndi said:

Homework Statement


A= {a1,...am}, B= {b1,...bn}. 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 and 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

Homework Equations

The Attempt at a Solution

[/B]The prove given is so short for me to understand anything.
It may help to take the principle you are applying and restate it using other variables.

Let D= {d1, ... , dm}, R= {r1, ... , rn}. If f: D→R is a function, then f(d1) can take anyone of the n values r1, ... , rn. Similarly for f(d2). Then there are nm such functions.​

Now apply this to the subset situation.
 
SammyS said:
Now apply this to the subset situation.
Sorry, that is the problem, I am unable to link the subset to the above mentioned situation.
 
rajeshmarndi said:
Sorry, that is the problem, I am unable to link the subset to the above mentioned situation.
What's stopping you? What part don't you understand? Simply saying "I don't get it" isn't very helpful.
 
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