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Question

Find the value of $$n$$ if $${}^{2n}C_3: {}^nC_3= 12: 1$$.


Solution

$$\dfrac { { ^{ 2n }{ C } }_{ 3 } }{ { { ^{ n }{ C } }_{ 3 } } } =\dfrac{12}{1}$$

$$\Rightarrow \dfrac { \dfrac { 2n! }{ \left( 2n-3 \right) !3! }  }{ \dfrac {n!}{ \left( n-3 \right) !3! }  } =12$$

$$\Rightarrow \dfrac { \left( n-3 \right) ! }{ n! } \times \dfrac { 2n! }{ \left( 2n-3 \right) ! } =12$$

$$\Rightarrow \dfrac { \left( 2n-2 \right)  }{ \left( n-2 \right) \left( n-1 \right) \left( n \right)  } \left( 2n-1 \right) \left( 2n \right) =12$$

$$\Rightarrow \dfrac{2\times 2\times \left(2n-1\right)}{n-2}=12$$
$$\Rightarrow  8n-4=12n-24$$
$$\Rightarrow 4n=20$$
$$\Rightarrow n=5.$$
Hence, the answer is $$5.$$

Mathematics

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