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Q.2.     Determine n if

            (ii)\; ^{2n}C_{3}:^{n}C_{3}=11:1

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Given that : (ii)\; ^{2n}C_{3}:^{n}C_{3}=11:1

            \Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{11}{1}

   \Rightarrow \, \, \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{11}{1}

 \Rightarrow \, \, {\frac{2n!(n-3)!}{(2n-3)!n!}}=\frac{11}{1}

  \Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times(n-2)\times (n-3)! }=\frac{11}{1}

\Rightarrow \, \, \frac{2n\times (2n-1)\times (2n-2)}{ n\times (n-1)\times(n-2) }=\frac{11}{1}

\Rightarrow \, \, \frac{2\times (2n-1)\times 2}{ (n-2) }=\frac{11}{1}

\Rightarrow \, \, \frac{4\times (2n-1)}{ (n-2) }=\frac{11}{1}

\Rightarrow \, \, 8n-4=11n-22

\Rightarrow \, \, 22-4=11n-8n

\Rightarrow \, \, 3n=18

\Rightarrow \, \, n=6

Thus the value of n=6

Posted by

seema garhwal

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