If the coefficients of 6th and 5th terms of expansion $(1+x)^n$ ;are in the ratio 7:5, then find the value of n.

Given,

$(1+x)^n$

General term

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{n}C_r}(1)^{n-r}x^r$

Given

$\dfrac{\ coeff \ of\ T_6}{coeff\ of\ T_5}=\dfrac{7}{5}$

$T_6=T_{5+1}={^{n}C_5}(1)^{n-5}x^5$

$T_5=T_{4+1}={^{n}C_4}(1)^{n-4}x^4$

$\dfrac{^{n}C_5}{^{n}C_4}=\dfrac{7}{5}$

$\dfrac{n!4!\times (n-4)!}{5!\times (n-5)!\times n!}=\dfrac{7}{5}$

$\dfrac{n!\times 4!\times (n-4)\times (n-5)!}{5!\times (n-5)!\times n!}=\dfrac{7}{5}$

$\dfrac{(n-4)}{5}=\dfrac{7}{5}$

$n=7+4$

$n=11$

Hence, correct option is A.