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Question Number 209281 by Tawa11 last updated on 06/Jul/24
Find the value of r, if ^(10) C_r = ^(10) C_(2r + 1)
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r},\:\mathrm{if}\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{r}} \:\:=\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{2r}\:\:+\:\:\mathrm{1}} \\ $$
Commented by klipto last updated on 06/Jul/24
^n C_r =^n C_(n−r) ^(10) C_r =^(10) C_(9−2r) ∴r=9−2r r+2r=9,r=3 or if r=2r+1,r=10−(2r+1),r=10−2r−1 r+2r=9,3r=9,r=3 klipto−quanta
$$\:\:\:\:\:\:\:^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{C}}_{\mathrm{r}} =^{\mathrm{n}} \boldsymbol{\mathrm{C}}_{\mathrm{n}−\mathrm{r}} \\ $$$$\:\:\:\:\:\:\:^{\mathrm{10}} \boldsymbol{\mathrm{C}}_{\boldsymbol{\mathrm{r}}} =^{\mathrm{10}} \boldsymbol{\mathrm{C}}_{\mathrm{9}−\mathrm{2}\boldsymbol{\mathrm{r}}} \\ $$$$\therefore\boldsymbol{\mathrm{r}}=\mathrm{9}−\mathrm{2}\boldsymbol{\mathrm{r}} \\ $$$$\boldsymbol{\mathrm{r}}+\mathrm{2}\boldsymbol{\mathrm{r}}=\mathrm{9},\boldsymbol{\mathrm{r}}=\mathrm{3} \\ $$$$\boldsymbol{\mathrm{or}} \\ $$$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{r}}=\mathrm{2}\boldsymbol{\mathrm{r}}+\mathrm{1},\boldsymbol{\mathrm{r}}=\mathrm{10}−\left(\mathrm{2}\boldsymbol{\mathrm{r}}+\mathrm{1}\right),\boldsymbol{\mathrm{r}}=\mathrm{10}−\mathrm{2}\boldsymbol{\mathrm{r}}−\mathrm{1} \\ $$$$\boldsymbol{\mathrm{r}}+\mathrm{2}\boldsymbol{\mathrm{r}}=\mathrm{9},\mathrm{3}\boldsymbol{\mathrm{r}}=\mathrm{9},\boldsymbol{\mathrm{r}}=\mathrm{3} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$
Commented by mr W last updated on 06/Jul/24
please use “answer” instead of “comment” when you want to answer a question.
$${please}\:{use}\:“{answer}''\:{instead}\:{of}\: \\ $$$$“{comment}''\:{when}\:{you}\:{want}\:{to}\:{answer} \\ $$$${a}\:{question}. \\ $$
Answered by A5T last updated on 06/Jul/24
((10!)/(r!(10−r)!))=((10!)/((10−2r−1)!(2r+1)!)) ⇒(10−2r−1)!(2r+1)!=r!(10−r)! 2r+1≤10⇒r≤4⇒r=3
$$\frac{\mathrm{10}!}{{r}!\left(\mathrm{10}−{r}\right)!}=\frac{\mathrm{10}!}{\left(\mathrm{10}−\mathrm{2}{r}−\mathrm{1}\right)!\left(\mathrm{2}{r}+\mathrm{1}\right)!} \\ $$$$\Rightarrow\left(\mathrm{10}−\mathrm{2}{r}−\mathrm{1}\right)!\left(\mathrm{2}{r}+\mathrm{1}\right)!={r}!\left(\mathrm{10}−{r}\right)! \\ $$$$\mathrm{2}{r}+\mathrm{1}\leqslant\mathrm{10}\Rightarrow{r}\leqslant\mathrm{4}\Rightarrow{r}=\mathrm{3} \\ $$
Commented by Tawa11 last updated on 06/Jul/24
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by A5T last updated on 06/Jul/24
r=2r+1 or r=10−(2r+1)⇒r=3
$${r}=\mathrm{2}{r}+\mathrm{1}\:{or}\:{r}=\mathrm{10}−\left(\mathrm{2}{r}+\mathrm{1}\right)\Rightarrow{r}=\mathrm{3} \\ $$

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