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Question Number 73029 by mathmax by abdo last updated on 05/Nov/19
prove that ∀(n,p,q)∈N^3 Σ_(k=0) ^n C_p ^k C_q ^(n−k) =C_(p+q) ^n conclude that Σ_(k=0) ^n (C_n ^k )^2 =C_(2n) ^n
$${prove}\:{that}\:\:\forall\left({n},{p},{q}\right)\in{N}^{\mathrm{3}} \:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{p}} ^{{k}} \:{C}_{{q}} ^{{n}−{k}} \:\:\:={C}_{{p}+{q}} ^{{n}} \\ $$$${conclude}\:{that}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \:={C}_{\mathrm{2}{n}} ^{{n}} \\ $$
Answered by mind is power last updated on 05/Nov/19
C_p ^k .C_q ^(n−k) =C_(p+q) ^n .... 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 ⇒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 =Σ_(k=0) ^n C_p ^k .C_q ^(n−k) 2nd we can mix A/and B we get a/box withe p+q object to pick n objdct=C_n ^(p+q) ⇒Σ_(k=0) ^n C_p ^k .C_q ^(n−k) =C_(p+q) ^n if p=q=n ⇒Σ_(k=0) ^n C_n ^k .C_n ^(n−k) =C_(2n) ^n C_n ^(n−k) =C_n ^k ⇒ Σ(C_n ^k )^2 =C_(2n) ^n
$$\mathrm{C}_{\mathrm{p}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{q}} ^{\mathrm{n}−\mathrm{k}} =\mathrm{C}_{\mathrm{p}+\mathrm{q}} ^{\mathrm{n}} …. \\ $$$$\mathrm{let}\:\mathrm{A}\:\mathrm{bee}\:\mathrm{a}\:\mathrm{box}\:\mathrm{withe}\:\mathrm{p}\:\mathrm{object} \\ $$$$\mathrm{let}\:\mathrm{B}\:\mathrm{bee}\:\mathrm{a}\:\mathrm{box}\:\mathrm{withe}\:\mathrm{q}\:\mathrm{objects} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{too}\:\mathrm{pick}\:\mathrm{n}\:\mathrm{object}\:\mathrm{in}\:\mathrm{the}\:\mathrm{too}\:\mathrm{box} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{k}+\mathrm{n}−\mathrm{k}\:\mathrm{pick}\:\mathrm{k}\:\mathrm{object}\:\mathrm{in}\:\mathrm{A}\:\mathrm{and}\:\mathrm{n}−\mathrm{k}\:\mathrm{in}\:\mathrm{B} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{p}} ^{\mathrm{k}} \:\mathrm{fir}\:\mathrm{A}\:\mathrm{and}\:\mathrm{C}_{\mathrm{q}} ^{\mathrm{n}−\mathrm{k}} \:\mathrm{in}\:\mathrm{B} \\ $$$$\mathrm{so}\:\mathrm{C}_{\mathrm{p}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{q}} ^{\mathrm{n}−\mathrm{k}} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{issues} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{p}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{q}} ^{\mathrm{n}−\mathrm{k}} \\ $$$$\mathrm{2nd}\:\mathrm{we}\:\mathrm{can}\:\mathrm{mix}\:\mathrm{A}/\mathrm{and}\:\mathrm{B}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}/\mathrm{box}\:\mathrm{withe}\:\mathrm{p}+\mathrm{q}\:\mathrm{object} \\ $$$$\mathrm{to}\:\mathrm{pick}\:\mathrm{n}\:\mathrm{objdct}=\mathrm{C}_{\mathrm{n}} ^{\mathrm{p}+\mathrm{q}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{p}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{q}} ^{\mathrm{n}−\mathrm{k}} =\mathrm{C}_{\mathrm{p}+\mathrm{q}} ^{\mathrm{n}} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{q}=\mathrm{n} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{n}} ^{\mathrm{n}−\mathrm{k}} =\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \\ $$$$\mathrm{C}_{\mathrm{n}} ^{\mathrm{n}−\mathrm{k}} =\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \Rightarrow \\ $$$$\Sigma\left(\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \right)^{\mathrm{2}} =\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
thankx sir.
$${thankx}\:{sir}. \\ $$
Commented by mind is power last updated on 06/Nov/19
y′re welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

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