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Question Number 73040 by mathmax by abdo last updated on 05/Nov/19

p r o v e t h a t \forall (n, p) \in N^{★} \times N

1) \sum_{k = 0}^{p} {(- 1)}^{k} C_{n}^{k} = {(- 1)}^{p} C_{n - 1}^{p}

2) \forall (p, q) \in N^{2} \sum_{k = 0}^{p} C_{p + q}^{k} C_{p + q - k}^{p - k} = 2^{p} C_{p + q}^{p}

Answered by mind is power last updated on 06/Nov/19

1) recursion

2 nd

lets A = [1,,, p + q]

in A we can get

C_{p + q}^{p} set withe p elements

the number of idont now the nam in inglish

^{″} A = {1, 2} / P (A) = {\emptyset, {1}, {2}, {1, 2}}^{″}

if A is card p \Rightarrow P (A) = 2^{p}

sl the number all partition of set withe p element is 2^{p} C_{p + q}^{p}

we can chose this element

we pick k element in p + q

and p - k in p + q - k

\Rightarrow= \underset{k = 0}{\sum^{p}} C_{p + q}^{k} C_{p + q - k}^{p - k} = 2^{p} C_{p + q}^{p}