Understanding a special combination

  • Thread starter Thread starter anigeo
  • Start date Start date
  • Tags Tags
    Combination
AI Thread Summary
The discussion centers on the equation nC0+nC1+nC2+.....+nCn=2^n, which represents the total number of combinations of n different items taken at least one at a time. Each of the n objects can be either accepted or rejected, leading to 2^n possible combinations. The significance of rejection is that it emphasizes the binary choice for each object, which is crucial for understanding the proof. The binomial theorem provides a simpler proof by expanding (a+b)^n and substituting a=b=1 to derive the result. Ultimately, the equation holds true only under the condition of two choices per object, reinforcing the concept of binary selection.
anigeo
Messages
83
Reaction score
0
nC0+nC1+nC2+.....+nCn=2^n

in the analytic proof for this my books say that it is the total number of combinations of n different things taken at least 1 at a time.
they say that each object can be dealt in 2 ways, either it can be accepted or it can be rejected.
hence n objects can be dealt in 2^n ways.
but how in selection how does the question of rejection come?what is the significance of this rejection?please explain.
 
Physics news on Phys.org
I can't understand your question. However a simple proof of the equation is by means of the binomial theorem. Expand (a+b)n and then let a=b=1 and you will get the result.
 
mathman said:
I can't understand your question. However a simple proof of the equation is by means of the binomial theorem. Expand (a+b)n and then let a=b=1 and you will get the result.

the binomial proof is done.this is the analytic proof in which they say that every object of n objects can be dealt in 2 ways,s objects can be dealt with in 2^2 ways,3 in 2^3 ways.this includes an acceptance of the object or its rejection.acceptance or selection is all right.how does the question of rejection arise?
 
It looks to be a matter of wording. Since every object can be dealt with in either of two ways, the two ways may be labeled "acceptance" or "rejection".
 
that's the thing, u got it.but how can u say that it can be dealt in only 2 ways?
 
anigeo said:
that's the thing, u got it.but how can u say that it can be dealt in only 2 ways?
That's what it is to prove the particular equation. If it's more than two ways, you have a different expression.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top