• ^n​C₇ = ⁿC₉, find the value of "n"
• ⁿC_{n-4 = 15, find "n"}

Dear Student,
Q1)

$\Rightarrow$ⁿC₇ = ⁿC₉ 
$\Rightarrow$$\frac{\mathrm{n}!}{7! (\mathrm{n}-7)!}=\frac{\mathrm{n}!}{9! (\mathrm{n}-9)!}$
$\Rightarrow$9! (n - 9)! = 7! (n - 7)!
$\Rightarrow$9 x 8 x 7! (n - 9)(n - 8)(n - 7)! = 7! (n - 7)!
$\Rightarrow$72 (n - 9) (n - 8) = 0
$\Rightarrow$(n - 9) (n - 8) = 0
$\Rightarrow$n = 8 , 9

Q2) 
$\Rightarrow$ⁿC_n-4 = 15
$\Rightarrow$$\frac{\mathrm{n}!}{(\mathrm{n}-4)! (\mathrm{n}-\left(\mathrm{n}-4\right))!}$ = 15
$\Rightarrow$$\frac{\mathrm{n}!}{(\mathrm{n}-4)! 4!}$ = 15
$\Rightarrow$$\frac{\mathrm{n}\left(\mathrm{n}-1\right)\left(\mathrm{n}-2\right)\left(\mathrm{n}-3\right)\left(\mathrm{n}-4\right)!}{\left(\mathrm{n}-4\right)! 4!}$ = 15

$\Rightarrow$n (n - 1)(n - 2)(n - 3) = 15 x 4!
$\Rightarrow$n (n - 1)(n - 2)(n - 3) = 15 x 24 
$\Rightarrow$n (n - 1)(n - 2)(n - 3) = 360
$\Rightarrow$n (n - 1)(n - 2)(n - 3) = 6 x 5 x 4 x 3
Therefore, n = 6

Regards