Sampling

The generalized differentiation is sampling. Let's say that John has 3 chickens and 2 ducks and he wants to remove 2 birds.

$$ { C \cdot C.C.D.D\over 2(C+D)   } = { C.C.C.D.D\over { 2(C) + C.D + 2(D) } } $$

$$ = { C.C.C.D.D\over { 2(C)} } + {C.C.C.D.D\over {C.D}} + { C.C.C.D.D\over {2(D)}} $$

$$ = {3\choose 2} .C.D.D + 3.2.C.C.D + {2\choose 2} .C.C.C  $$

$$ = 3.C.D.D + 6.C.C.D + 1.C.C.C $$

One may conclude, for example, that the probability for John to remove two birds of different kind is 6/10.

in the above I have noted the differentiation by a ratio line and a set of 3 chickens by 3(C).

Eventually, we have reached in our research the formula of binomial coefficient.

{ X.X.....X.X \over \{ X,X,...,X \}   }={   X^n \over k(X)  } = {n\choose k}.X^{n-k}

k-lists with and without repetition

{ X.X.....X.X \over \{ X,X,...,X \}   }={   X^n \over { \lbrace X \rbrace }^k  } = {n\choose k}.X^{n-k}

{ X.X.....X.X \over  X.....X  }={  X^n \over X^k } = {n!\over (n-k)!}.X^{n-k}

of course,

{ X^n \over  X^n}  = n!.1