prove-that-n-p-q-N-3-k-0-n-C-p-k-C-q-n-k-C-p-q-n-conclude-that-k-0-n-C-n-k-2-C-2n-n-

Question Number 73029 by mathmax by abdo last updated on 05/Nov/19

p r o v e t h a t \forall (n, p, q) \in N^{3} \sum_{k = 0}^{n} C_{p}^{k} C_{q}^{n - k} = C_{p + q}^{n}

c o n c l u d e t h a t \sum_{k = 0}^{n} {(C_{n}^{k})}^{2} = C_{2 n}^{n}

Answered by mind is power last updated on 05/Nov/19

C_{p}^{k} . C_{q}^{n - k} = C_{p + q}^{n} \dots .

let A bee a box withe p object

let B bee a box withe q objects

we want too pick n object in the too box

\Rightarrow n = k + n - k pick k object in A and n - k in B

we have C_{p}^{k} fir A and C_{q}^{n - k} in B

so C_{p}^{k} . C_{q}^{n - k}

number of all issues

= \underset{k = 0}{\sum^{n}} C_{p}^{k} . C_{q}^{n - k}

2 nd we can mix A / and B we get a / box withe p + q object

to pick n objdct = C_{n}^{p + q}

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

if p = q = n

\Rightarrow \underset{k = 0}{\sum^{n}} C_{n}^{k} . C_{n}^{n - k} = C_{2 n}^{n}

C_{n}^{n - k} = C_{n}^{k} \Rightarrow

Σ {(C_{n}^{k})}^{2} = C_{2 n}^{n}

Commented by mathmax by abdo last updated on 05/Nov/19

t h a n k x s i r .

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

y^{'} re welcom