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Question Number 74000 by mhmd last updated on 17/Nov/19

if p_r ^n =840 , c_r ^n =35 find value of r?  pleas sir help me?

$${if}\:{p}_{{r}} ^{{n}} =\mathrm{840}\:,\:{c}_{{r}} ^{{n}} =\mathrm{35}\:{find}\:{value}\:{of}\:{r}? \\ $$$${pleas}\:{sir}\:{help}\:{me}? \\ $$

Answered by MJS last updated on 17/Nov/19

C_r ^n =((n!)/(r!(n−r)!))  P_r ^( n) =((n!)/((n−r)!))=C_r ^n ×r!  C_r ^n =35  P_r ^( n) =840  35r!=840  r!=24  r=4  ⇒  ((n!)/((n−4)!))=840  n(n−1)(n−2)(n−3)=840  n^4 −6n^3 +11n^2 −6n−840=0  (n−7)(n+4)(n^2 −3n+30)=0  n≥0 ⇒ n=7

$${C}_{{r}} ^{{n}} =\frac{{n}!}{{r}!\left({n}−{r}\right)!} \\ $$$${P}_{{r}} ^{\:{n}} =\frac{{n}!}{\left({n}−{r}\right)!}={C}_{{r}} ^{{n}} ×{r}! \\ $$$${C}_{{r}} ^{{n}} =\mathrm{35} \\ $$$${P}_{{r}} ^{\:{n}} =\mathrm{840} \\ $$$$\mathrm{35}{r}!=\mathrm{840} \\ $$$${r}!=\mathrm{24} \\ $$$${r}=\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\frac{{n}!}{\left({n}−\mathrm{4}\right)!}=\mathrm{840} \\ $$$${n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)\left({n}−\mathrm{3}\right)=\mathrm{840} \\ $$$${n}^{\mathrm{4}} −\mathrm{6}{n}^{\mathrm{3}} +\mathrm{11}{n}^{\mathrm{2}} −\mathrm{6}{n}−\mathrm{840}=\mathrm{0} \\ $$$$\left({n}−\mathrm{7}\right)\left({n}+\mathrm{4}\right)\left({n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{30}\right)=\mathrm{0} \\ $$$${n}\geqslant\mathrm{0}\:\Rightarrow\:{n}=\mathrm{7} \\ $$

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