Identity of Vandermonde

In Mathematical Combinative S, the identity of Vandermonde , named according to Alexandre-Theophilus Vandermonde, affirms that

${n+m \ choose R} = \ sum_ {k=0} ^r {N \ choose K} {m \ choose r-k}.$

Proof

Algebraic

It can be shown in an algebraic way.

The product of two polynomials to a variable is given by:

$\ left (\ sum_ {i=0} ^n a_ix^i \ right) \ left (\ sum_ {i=0} ^m b_ix^i \ right) = \ sum_ {k=0} ^ {m+n} \ left (\ sum_ {j=0} ^k a_jb_ {k-j} \ right) x^k \ quad \ text {with} m, N \ in \ mathbb {NR}$

By the theorem of the binomial theorem, we know that

$(1+x) ^ {m+n} = \ sum_ {r=0} ^ {m+n} \ binom {m+n} {R} x^r$

We can transform this equality into product of polynomials.

$\ begin {array} {L}$

\ sum_ {r=0} ^ {m+n} \ binom {m+n} {R} x^r & = (1+x) ^ {m+n} \ \ & = (1+x) ^ {N} (1+x) ^ {m} \ \ & = \ left (\ sum_ {i=0} ^ {N} \ binom {N} {I} x^i \ right) \ left (\ sum_ {j=0} ^ {m} \ binom {m} {J} x^j \ right) \end{array}

By using the equation of the higher product of polynomials and by simplifying the result, the identity of Vandermonde appears:

\ begin {array} {L} \ sum_ {r=0} ^ {m+n} \ binom {m+n} {R} x^r & = \ sum_ {r=0} ^ {m+n} \ left (\ sum_ {k=0} ^ {R} \ binom {N} {K} \ binom {m} {r-k} \ right) x^r \ \ \ binom {m+n} {R} x^r & = \ left (\ sum_ {k=0} ^ {R} \ binom {N} {K} \ binom {m} {r-k} \ right) x^r \ \ \ binom {n+m} {R} & = \ sum_ {k=0} ^ {R} \ binom {N} {K} \ binom {m} {r-k} \end{array}

Bijective

A bijective Preuve is also possible. Cash let us suppose that a parliamentary committee is composed members of two political parties only, one N members, the “green”, the other m members, the “yellows”. How much can one form committees of size R starting from these two parties? The answer is of course

${n+m \ choose R}.$

This value is also given by the sum of all the values of K of the number of committees made up of K green and R − K yellow.

Hypergeometric probability distribution

When the two sides of this identity are divided by the expression with the left, then the terms obtained can be interpreted like probabilities, which are given by the hypergeometric Distribution. It is the probability of drawing from the red balls in R pullings without handing-over of a ballot box containing red N balls and m balls blue. For example, let us suppose that a person is responsible to randomly create a committee of R drawn members among green N and m yellow. Then which is the probability that there is exactly K green in the committee? The answer is in this distribution.

Identity of Chu-Vandermonde

The identity of Chu-Vandermonde generalizes it for not-whole values:

${s+t \ choose N} = \ sum_ {k=0} ^ \ infty {S \ choose K} {T \ choose n-k}$

It is true for all complex numbers S and T .

It is a special case of the hypergeometric Théorème of Gauss which affirms that

$\;_2F_1 (has, B; C; 1) = \ frac {\ Gamma (c) \ Gamma (c-a-b)}{\ Gamma (Ca) \ Gamma (c-b)}$

where $\; _2F_1$ is the hypergeometric Série and $\ Gamma (n+1) =n!$ is the Fonction gamma. It is enough to apply has = − N and the identity

{N \ choose K} = (- 1) ^k {k-n-1 \ choose K}

on several occasions.

External bond

BinomialCoefficients contains some demonstrations of the identity of Vandermonde

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