The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is ________.

A : number of ways that all boys 
sit together = 5! × 5! 
B : number of ways if no 2 boys 
sit together = 4! × 5! 
$\mathrm{A}\cap \mathrm{B}=\mathrm{\varphi }$
Required no. of ways = 5! × 5! + 4! × 5! = 17280