Prove that the number of ways in which $8$ different flowers can be strung to form a garland so that $4$ particular flowers are never separated is $\frac{1}{2}4!4!$

Tie 4 flowers so that we have now in all five (4 different and- one bunch). These five can be arranged in a garland in $\frac{1}{2}$ i.e. (n - 1)! i.e. $\frac{1}{2}$ (4!) ways. The bunch can be arranged in 4! ways. Hence by fundamental theorem total is $\frac{1}{2}(4!.4!)$