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 ________.
Answer (integer)
17280
Solution
<p>A : number of ways that all boys sit together $=5!\times 5!$</p>
<p>B : number of ways if no 2 boys</p>
<p>sit together $=4!\times 5!$</p>
<p>$\mathrm{A} \cap \mathrm{B}=\phi$</p>
<p>Required no. of ways $=5!\times 5!+4!\times 5!=17280$</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Permutations
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