Given,
\( \dfrac{^{2n}C_3}{^nC_3}=\dfrac{12}{1}\)
\( \dfrac{(2n)!}{3!(2n-3)!}\times \dfrac{3!(n-3)!}{n!}=\dfrac{12}{1}\)
\( \dfrac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}\)
\( \times \dfrac{(n-3)!}{n(n-1)(n-2)(n-3)!}=12\)
= \(\dfrac{2(2n-1)(2n-2)}{(n-1)(n-2)}=12\)
= \( \dfrac{4(2n-1)(n-1)}{(n-1)(n-2)}=12\)
= \( \dfrac{(2n-1)}{(n-2)}=12\)
= 2n -1=3(n - 2)
= 2n -1=3n -6
= 3n -2n = -1+6
= n =5
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