Total balls=N
blue balls= n
red balls = N-n
probability of n blue balls in first n draw
$\frac{^{n}C_{n}}{^{N}C_{n}}{}$
probailty of (N-n) red balls in first (N-n) draw
$\frac{^{(N-n)}C_{(N-n)}}{^{N}C_{(N-n)}}$
according to question
$\frac{^{n}C_{n}}{^{N}C_{n}}{}$=$\frac{^{(N-n)}C_{(N-n)}}{^{N}C_{(N-n)}}$
$\implies\frac{1}{^{N}C_{(n)}}=\frac{1}{^{N}C_{(N-n)}}$
$\implies\frac{(N-n)! n!}{N!}=\frac{(N-(N-n))!(N-n)!}{N!}$
$\implies\frac{(N-n)!n!}{N!}=\frac{(N-n)!n!}{N!}$
LHS=RHS
Second method
if total balls =5
blue balls =2
then red balls = 5-2=3
probabilty of picking all blue balls in first 2 draw =$\frac{^{2}C_{2}}{^{5}C_{2}}$
probabilty of picking all red balls in first 3 draw = $\frac{^{3}C_{3}}{^{5}C_{3}}$
according to question
$\implies\frac{^{2}C_{2}}{^{5}C_{2}}=\frac{^{3}C_{3}}{^{5}C_{3}}$
$\implies\frac{1}{10}=\frac{1}{10}$
LHS=RHS