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…
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 ________ -
Solution
A : number of ways that all boys sit together $=5!\times 5!$ B : number of ways if no 2 boys sit together $=4!\times 5$ ! $\mathrm{A} \cap \mathrm{B}=\phi$ Required no. of ways $=5!\times 5!+4!\times 5!=17280$