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

$$provethat\mathrm{\forall }(n,p,q)\in N^3\sum _{k=0}^n{C}_p^k{C}_q^{n-k}=C_{p+q}^n$$$$concludethat\sum _{k=0}^n{\left(C_n^k\right)}^2=C_{2n}^n$$

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

$${\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}.{\mathrm{C}}_{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}={\mathrm{C}}_{\mathrm{p}+\mathrm{q}}^{\mathrm{n}}....$$$$\mathrm{let}\mathrm{A}\mathrm{bee}\mathrm{a}\mathrm{box}\mathrm{withe}\mathrm{p}\mathrm{object}$$$$\mathrm{let}\mathrm{B}\mathrm{bee}\mathrm{a}\mathrm{box}\mathrm{withe}\mathrm{q}\mathrm{objects}$$$$\mathrm{we}\mathrm{want}\mathrm{too}\mathrm{pick}\mathrm{n}\mathrm{object}\mathrm{in}\mathrm{the}\mathrm{too}\mathrm{box}$$$$\Rightarrow \mathrm{n}=\mathrm{k}+\mathrm{n}-\mathrm{k}\mathrm{pick}\mathrm{k}\mathrm{object}\mathrm{in}\mathrm{A}\mathrm{and}\mathrm{n}-\mathrm{k}\mathrm{in}\mathrm{B}$$$$\mathrm{we}\mathrm{have}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}\mathrm{fir}\mathrm{A}\mathrm{and}{\mathrm{C}}_{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}\mathrm{in}\mathrm{B}$$$$\mathrm{so}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}.{\mathrm{C}}_{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}$$$$\mathrm{number}\mathrm{of}\mathrm{all}\mathrm{issues}$$$$=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}.{\mathrm{C}}_{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}$$$$2\mathrm{nd}\mathrm{we}\mathrm{can}\mathrm{mix}\mathrm{A}/\mathrm{and}\mathrm{B}\mathrm{we}\mathrm{get}\mathrm{a}/\mathrm{box}\mathrm{withe}\mathrm{p}+\mathrm{q}\mathrm{object}$$$$\mathrm{to}\mathrm{pick}\mathrm{n}\mathrm{objdct}={\mathrm{C}}_{\mathrm{n}}^{\mathrm{p}+\mathrm{q}}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{p}}^{\mathrm{k}}.{\mathrm{C}}_{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}={\mathrm{C}}_{\mathrm{p}+\mathrm{q}}^{\mathrm{n}}$$$$\mathrm{if}\mathrm{p}=\mathrm{q}=\mathrm{n}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}.{\mathrm{C}}_{\mathrm{n}}^{\mathrm{n}-\mathrm{k}}={\mathrm{C}}_{2\mathrm{n}}^{\mathrm{n}}$$$${\mathrm{C}}_{\mathrm{n}}^{\mathrm{n}-\mathrm{k}}={\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\Rightarrow$$$$\mathrm{\Sigma }{\left({\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}\right)}^2={\mathrm{C}}_{2\mathrm{n}}^{\mathrm{n}}$$

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

$$thankxsir.$$

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

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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