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Question Number 185642 by Mingma last updated on 24/Jan/23

Commented by mr W last updated on 24/Jan/23

=C_(4+1) ^(26+1) =C_5 ^(27) =80730

$$={C}_{\mathrm{4}+\mathrm{1}} ^{\mathrm{26}+\mathrm{1}} ={C}_{\mathrm{5}} ^{\mathrm{27}} =\mathrm{80730} \\ $$

Answered by cortano1 last updated on 25/Jan/23

Σ_(k=4) ^n C_4 ^k = (1/(4!))((((n−3)(n−2)(n−1)n(n+1))/5))

$$\:\underset{{k}=\mathrm{4}} {\overset{{n}} {\sum}}\:{C}_{\mathrm{4}} ^{{k}} \:=\:\frac{\mathrm{1}}{\mathrm{4}!}\left(\frac{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)}{\mathrm{5}}\right)\: \\ $$

Commented by mr W last updated on 25/Jan/23

Σ_(k=4) ^n C_4 ^k = (1/(4!))((((n−3)(n−2)(n−1)n(n+1))/5)) =(((n+1)!)/(5!(n−4)!))=C_5 ^(n+1) generally Σ_(k=r) ^n ((k),(r) ) = (((n+1)),((r+1)) ) or Σ_(k=r) ^n C_r ^k =C_(r+1) ^(n+1) this is the Hockey−stick identity.

$$\:\underset{{k}=\mathrm{4}} {\overset{{n}} {\sum}}\:{C}_{\mathrm{4}} ^{{k}} \:=\:\frac{\mathrm{1}}{\mathrm{4}!}\left(\frac{\left({n}−\mathrm{3}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)}{\mathrm{5}}\right) \\ $$$$\:=\frac{\left({n}+\mathrm{1}\right)!}{\mathrm{5}!\left({n}−\mathrm{4}\right)!}={C}_{\mathrm{5}} ^{{n}+\mathrm{1}} \\ $$$${generally} \\ $$$$\underset{{k}={r}} {\overset{{n}} {\sum}}\begin{pmatrix}{{k}}\\{{r}}\end{pmatrix}\:=\begin{pmatrix}{{n}+\mathrm{1}}\\{{r}+\mathrm{1}}\end{pmatrix}\:{or}\:\underset{{k}={r}} {\overset{{n}} {\sum}}{C}_{{r}} ^{{k}} ={C}_{{r}+\mathrm{1}} ^{{n}+\mathrm{1}} \\ $$$${this}\:{is}\:{the}\:{Hockey}−{stick}\:{identity}. \\ $$

Commented by mr W last updated on 25/Jan/23

proof see Q185659

$${proof}\:{see}\:{Q}\mathrm{185659} \\ $$

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