Question

There are 15 people to be seated around a round table. In how many ways we can arrange them such that A, B, C all three are never together

• A. $12!\times 3!$
• B. $12!\times 176$
• C. $12!\times 182$
• D. $12!\times 142$

Answer 1

The correct option is B $12!\times 176$

Total no. of required permutations
(Total circular permutations) - (Total circular permutations when all 3 are together.)

$=14!-12!\times 3!=12![14\times 13-6]$
$=12![182-6]=12!\times \left[176\right]$